CBSE Class 12

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 1 The Solid State. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 1 Important Extra Questions The Solid State

The Solid State Important Extra Questions Very Short Answer Type

Question 1.
A metal crystallises in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’? (CBSE Sample Paper 2017-18)
Answer:
r = \(\frac{\sqrt{3}}{4}\)a

Question 2.
If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’. What is the relation between ‘r’ and ‘R’? (CBSE Sample Paper 2017-18)
Answer:
r = 0.414 R

Question 3.
How many octahedral voids are there in 1 mole of a compound having cubic closed packed structure? (CBSE Sample Paper 2007)
Answer:
4 mole

Question 4.
Write a feature which will distinguish a metallic solid from an ionic solid.
Answer:
Metallic solids are good conductors of heat and electricity whereas ionic solids are insulators in solid state but conductors in molten state and in aqueous solution.

Question 5.
Write a distinguishing feature of metallic solids.
Answer:
Metallic solids are good conductors of heat and electricity.

Question 6.
KF has ccp structure. Calculate the radius of the unit cell if the side of the cube or edge length is 400 pm. How many F ions and octahedral voids are there in the unit cell? (CBSE Sample Paper 2011)
Answer:
For ccp lattice,
r = \(\frac{a}{2 \sqrt{2}}=\frac{400}{2 \times 1.414}\) = 141.4 pm
There are four (4) F ions and four (4) octahedral voids per unit cell.

Question 7.
Which point defect in crystals lowers the density of a crystal?
Answer:
Schottky defect

Question 8.
What type of interactions hold the molecules together in a polar molecular solid?
Answer:
Dipole-dipole forces

Question 9.
Which stoichiometric defect in crystals increases the density of a solid? (CBSE Delhi 2012)
Answer:
Interstitial defect

Question 10.
What type of semiconductor is obtained when silicon is doped with arsenic? (C.B.S.E. 2010)
Answer:
n-type semiconductor

Question 11.
Define forbidden zone of an insulator. (C.B.S.E. Delhi 2008)
Answer:
The large energy gap between the filled valence band and the empty conduction band in an insulator is called forbidden zone.

Question 12.
How do metallic and ionic substances differ in conducting electricity? (C.B.S.E. 2009)
Answer:
Metallic solids conduct electricity in the solid state whereas ionic substances conduct electricity in molten state or in solution, or metallic substances conduct electricity through electrons while ionic substances conduct electricity through ions.

Question 13.
What is the coordination number of each type of ions in a rock salt type crystal structure? (C.B.S.E. Delhi 2008)
Answer:
C.N. of Na⁺ = 6, C.N. of Cl^– = 6.

Question 14.
What is meant by ‘doping’ in a semiconductor? (C.B.S.E. Delhi 2012)
Answer:
The process of introduction of small amounts of impurities in the lattice of the crystal is called doping.

Question 15.
How may the conductivity of an intrinsic semiconductor be increased? (C.B.S.E. Delhi 2012)
Answer:
The conductivity of an intrinsic semiconductor may be increased by adding an appropriate amount of suitable impurity. This process is called doping.

Question 16.
How many atoms constitute one unit cell of a face-centred cubic crystal? (C.B.S.E. Delhi 2013)
Answer:
Four

Question 17.
What type of magnetism is shown by a substance if magnetic moments of domains are arranged in the same direction? (C.B.S.E. Delhi 2016)
Answer:
Ferromagnetism

Question 18.
Out of NaCI and AgCI, which one shows Frenkel defect and why? (C.B.S.E. Delhi 2019)
Answer:
AgCI shows Frenkel defect. This is because of small size of Ag⁺ ions; these can fit into interstitial sites of Cl^– ions.

Question 19.
What type of stoichiometric defect is shown by ZnS and why? (C.B.S.E. Delhi 2019)
Answer:
Frenkel defect. This is caused due to the large difference in size of ions.

Question 20.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

The Solid State Important Extra Questions Short Answer Type

Question 1.
Calculate the number of unit cells in 8.1 g of aluminium if it crystallises in a face-centred cubic (fcc) structure. (Atomic mass of Al = 27 g mol^-1).
Answer:
Moles of aluminium = \(\frac{8.1}{27}\) mol
No. of atoms of Al in 8.1 g = \(\frac{8.1}{27}\) × 6.022 × 10²³
No. of atoms in fee unit cell = 4
No. of unit cells = \(\frac{8.1}{27} \times \frac{6.022 \times 10^{23}}{4}\)
= 4.5 × 10²²

Question 2.
Tungsten crystallises in body centred cubic unit cell. If edge of the unit cell is 316.5 pm, what is the radius of the tungsten atom? (CBSE Sample Paper 2017-18)
Answer:
If a is the edge length of bcc unit cell, then radius of an atom,
r = \(\frac{\sqrt{3}}{4}\)a
Here, a = 316.5 pm
∴ r = \(\frac{\sqrt{3}}{4}\) × 316.5 pm = 137.04 pm

Question 3.
Silver metal crystallises with a face centred cubic lattice. The length of the unit cell is found to be 4.077 × 10^-8 cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108u, N_A = 6.02 × 10²³). (C.B.S.E. Sample Paper 2012)
Answer:
Edge length of unit cell, a = 4.077 × 10^-8 cm For fcc lattice, radius of an atom is related to edge length, a as:

Question 4.
Analysis shows that FeO has a non- stoichiometric composition with molecular formula Fe_0.950. Give reason.
Answer:
It shows metal deficiency defect. In FeO, some Fe²⁺ ions are replaced by Fe³⁺ ions. Three Fe²⁺ ions are replaced by two Fe³⁺ ions to maintain electrical neutrality.

Question 5.
Following is the schematic alignment of magnetic moments:

Identify the type of magnetism. What happens when these substances are heated? (CBSE Sample Paper 2017-18)
Answer:
Ferrimagnetism. These substances lose ferrimagnetism on heating and become paramagnetic.

Question 6.
Identify the type of defect shown in the following figure:
What type of substances show this defect? (CBSE Sample Paper 2017-18)

Answer:
Schottky defect.
This type of defect is shown by ionic compounds in which
(i) the ions have high coordination number and
(ii) ions (cations and anions) are of almost similar sizes.

Question 7.
Classify each of the following as being either a p-type or n-type semiconductor:
(i) Ge doped with In.
(ii) Si doped with B. (CBSE Sample Paper 2011)
Answer:
(i) Ge belongs to group 14 and In belongs to group 13.
Therefore, an electron deficient hole is created and it is a p-type semiconductor.

(ii) B belongs to group 13 and Si belongs to group 14. Therefore, an electron deficient hole is created and it is a p-type semiconductor.

Question 8.
Give reason:
(a) Why is Frenkel defect found in AgCI?
(b) What is the difference between phosphorus doped and gallium doped silicon semiconductors? (CBSE Sample Paper 2011)
Answer:
(a) Due to small size of Ag⁺ ion, it can fit into interstitial sites.
(b) Phosphorus doped silicon is n-type semiconductor while gallium doped silicon is p-type semiconductor.

Question 9.
Why does LiCl acquire pink colour when heated in Li vapours? (CBSE Sample Paper 2011)
Answer:
On heating LiCl in Li vapours, the excess of Li atoms deposit on the surface of the crystal. The CT ions diffuse to the surface of the crystal and combine with Li atoms to form LiCl. The electrons produced by ionisation of Li atoms diffuse into the crystal and get trapped at anion vacancies called F-centres. These absorb energy from visible light and radiate pink colour.

Question 10.
Write a feature which will distinguish a metallic solid from an ionic solid. (C.B.S.E. Delhi 2010, 2012)
Answer:
Metallic solids are good conductors of heat and electricity whereas ionic solids are insulators in solid state but conductors in molten state and in aqueous solution.

Question 11.
Crystalline solids are anisotropic in nature. What does this statement mean? (C.B.S.E. Delhi 2011)
Answer:
This means that crystalline solids have different physical properties such as electrical resistance or refractive index in different directions. This is because of different arrangement of particles in different directions.

Question 12.
What are n-type semiconductors? (C.6.S.E. Delhi 2012)
Answer:
These are the semiconductors in which the current is carried by the electrons in the normal way. For example, germanium doped with impurity containing five valence electrons (e.g., P).

Question 13.
What is the formula of a compound in which the element Y forms hep lattice and atoms of X occupy 2/3rd of tetrahedral voids? (C.B.S.E. 2015)
Answer:
Atoms Y adopt hep arrangement and there are two tetrahedral sites per atom of Y. Since 2/3rd tetrahedral sites are occupied by atoms of X, then for each atom of Y, the number of X atoms will be = 2 × \(\frac{2}{3}=\frac{4}{3}\)
Formula of compound – X_4/3Y or X₄Y₃

Question 14.
Define the following terms in relation to crystalline solids:
(i) Unit cell
(ii) Coordination number
Give one example in each case. (C.B.S.E. 2011)
Answer:
(i) Unit cell is the smallest portion of a crystal lattice which when repeated over and over again in different directions results in the entire lattice. For example, simple cubic unit cell.

(ii) The number of nearest neighbours in a packing is called coordination number. For example, in a body centred cubic structure, the coordination number is eight.

Question 15.
Account for the following:
(i) Schottky defects lower the density of related solids.
(ii) Conductivity of silicon increases on doping it with phosphorus. (C.B.S.E. 2013)
Answer:
(i) In Schottky defect, there are holes due to missing cations and anions. Due to the presence of holes in solid, the density decreases.

(ii) Pure silicon has a network lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus having five valence electrons, the impurity leads to excess of electrons after forming four covalent bonds like silicon. The extra electrons serve to conductivity and therefore, the conductivity of silicon doped with phosphorus increases.

Question 16.
(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
Answer:
Metal excess defect due to anion vacancies filled by free electrons i.e. F-centres.

(ii) What type of stoichiometric defect is shown by NaCI?
Answer:
Schottky defect.

OR

How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
Answer:
Tetrahedral void is surrounded by 4 constituent particles (atoms, molecules or ions). Octahedral void is surrounded by 6 constituent particles (atoms, molecules or ions).

(ii) Crystal lattice and unit cell? (C.B.S.E. 2014)
Answer:
A regular three dimensional arrangement of points in space is called crystal lattice.
The smallest repeating pattern in crystal lattice which when repeated in three dimensional space gives the entire lattice is called the unit cell.

Question 17.
The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca²⁺ ions and 8 F^– ions and that Ca²⁺ ions are arranged in a fcc lattice. The fluoride ions fill all the tetrahedral holes in the face centred cubic lattice of Ca²⁺ ions. The edge of the unit cell is 5.46 × 10^-8 cm in length. The density of the solid is 3.18 g cm^-3. Use this information to calculate Avogadro’s number. (Molar mass of CaF₂ = 78.08 g mol”1) (C.B.S.E. Delhi 2010)
Answer:
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
d = 3.18 g cm^-3, M = 78.08 g mol^-1
a = 5.46 × 10^-8 cm
Since the lattice is fcc, Z = 4
Substituting the values,

Question 18.
The density of copper metal is 8.95 g cm^-3. If the radius of copper atom be 127.8 pm, is the copper unit cell simple cubic, body centred cubic or face centred cubic? (C.B.S.E. 2010, C.B.S.E. Delhi 2010) ZxM
Answer:
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
Assuming fcc lattice for copper
a = 2√2 r
= 2 × 1.414 × 127.8 × 10^-10 cm
∴ a³ = (2 × 1.414 × 127.8 × 10^-10 cm)³
= 4.721 × 10^-23 cm³
∴ d = \(\frac{4 \times 63.54}{4.721 \times 10^{-23} \times 6.02 \times 10^{23}}\)
= 8.94 g cm^-3
Since the density is same as observed, the lattice is fcc lattice.

Question 19.
An element with density 10 g cm^-3 forms a cubic unit cell with edge length of 3 × 10^-8 cm. What is the nature of the cubic unit cell if the atomic mass of the element is 81 g mol^-1? (C.B.S.E. 2015)
Answer:
Density, d = \(\frac{z \times M}{a^{3} \times N_{A}}\)
d = 10 g cm^-3, M = 81 g mol^-1
a³ = (3 × 10^-8 cm)³
= 27 × 10^-24 cm³
N_A = 6.022 × 10²³
Therefore, 10 = \(\frac{Z \times 81}{\left(27 \times 10^{-24}\right) \times 6.022 \times 10^{23}}\)
Z = \(\frac{10 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{81}\)
= 2.007
Nature of cubic unit cell = bcc

Question 20.
Aluminium crystallises in a fcc structure. Atomic radius of the metal is 125 pm. What is the length of the side of unit cell of the metal? (C.B.S.E. Delhi 2019C)
Answer:
For fcc,
r = \(\frac{a}{2 \sqrt{2}}\)
a = 2r × √2 = 2 × 125 pm × 1.414 = 353.5 pm

Question 21.
Answer the following:
(a) What is the formula of a compound in which element Y forms ccp lattice and atom X occupy \(\frac{1}{3}\)rd of tetrahedral voids?
(b) What type of non-stoichiometric point defect leads to colour in alkali metal halides? (C.B.S.E. Delhi 2019C)
Answer:
(a) X₂Y₃
(b) Metal excess defect due to anionic vacancies / F-centres.

Question 22.
(a) Atoms of element B form hep lattice and those of the element A occupy \(\frac{2}{3}\)rd of octahedral voids. What is the formula of the compound formed by the elements A and B?
(b) What type of stoichiometric defect is shown by ZnS and why? (C.B.S.E. Delhi 2019C)
Answer:
(a) A₂B₃
(b) Frenkel defect, due to small size of Zn²⁺ ion.

Question 23.
What happens when AgCl is doped with CdCl₂? What is the name of this defect?
OR
What type of defect is shown by NaCl in
(a) stoichiometric defects, and
(b) non-stoichiometric defects? (C.B.S.E. Delhi 2019C)
Answer:
When CdCl₂ is added to AgCl, it introduces impurity defect. The addition of one Cd²⁺ ion will replace two Ag⁺ ions to maintain electrical neutrality. One of the position of Ag⁺ ions will be occupied by Cd²⁺ ion and other will be left as a hole. Thus, cationic vacancies are produced. This is called impurity defect.
OR
(a) In stoichiometric defects: Schottky defect.
(b) In non-stoichiometric defects: Metal excess due to anionic vacancies.

The Solid State Important Extra Questions Long Answer Type

Question 1.
Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s number, N_A = 6.02 × 10²³ mol^-1). (CBSE 2012)
Answer:
For a fcc unit cell, edge length (a) is related to radius of atom as:
r = \(\frac{a}{2 \sqrt{2}}\)
or a = 2√2.r = 2 × 1.414 × 127.8
= 361.42 pm
or = 361.42 × 10^-10 cm
Since the lattice is fcc, the number of copper atoms per unit cell, Z = 4.
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
M = 63.55 u, N_A = 6.02 × 10²³
∴ d = \(\frac{4 \times 63.55}{\left(361.42 \times 10^{-10}\right)^{3} \times\left(6.02 \times 10^{23}\right)}\)
= 8.94 g cm^-3

Question 2.
An element crystallises in a fcc lattice with cell edge of 250 pm. Calculate its density if 300 g of this element contain 2 × 10²⁴ atoms.
Answer:
Length of edge, a = 250 pm = 250 × 10^-12 m
= 250 × 10^-10 cm
Volume of unit cell = (250 × 10^-10 cm)³
= 15.625 × 10^-24 cm³
Mass of unit cell = No. of atoms in unit cell × Mass of each atom
Since the element has fcc arrangement, the number of atoms per unit cell, Z = 4
Mass of an atom = \(\frac{300}{2 \times 10^{24}}\) g
∴ Mass of unit cell = \(\frac{300}{2 \times 10^{24}}\) x 4
= 6.0 × 10^-22 g

Question 3.
An element with molar mass 27 g mol^-1 forms a cubic unit cell with edge length 4.05 × 10^-8 cm. If its density is 2.7 g cm^-3, what is the nature of the unit cell?
Answer:
Density of unit cell, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)
d = 2.7 g cm^-3, a = 4.05 × 10^-8 cm,
M = 27 g mol^-1, NA = 6.022 × 10²³

= 4.0
Since Z = 4, the unit cell is face centred cubic (fcc) unit cell.

Question 4.
An element with density 11.2 g cm^-3 forms a fee lattice with edge length of 4 × 10^-8 cm. Calculate the atomic mass of the element.
Answer:
Edge length of the unit cell
a = 4 × 10^-8 cm
Density = 11.2 g cm^-3
No. of atoms per unit cell in fcc lattice, Z = 4
Density, d = \(\frac{z \times M}{a^{3} \times N_{A}}\)
11.2 g cm^-3 = \(\frac{4 \times M}{\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)}\)
or M = \(\frac{\left(11.2 \mathrm{~g} \mathrm{~cm}^{-3}\right) \times\left(4 \times 10^{-8} \mathrm{~cm}\right)^{3} \times\left(6.02 \times 10^{23} \mathrm{~mol}^{-1}\right)}{4}\)
= 107.9 g mol^-1
Atomic mass of element = 107.9 u.

Question 5.
The density of lead is 11.35 g cm^-3 and the metal crystallises with fcc unit cell. Estimate the radius of lead atom. (At. Mass of lead = 207 g mol^-1 and N_A = 6.02 × 10²³ mol^-1) (CBSE Delhi 2011)
Answer:
Let length of edge = a cm
Density = 11.35 g cm^-3
No. of atoms per unit cell in fcc Lattice = 4
Atomic mass, M = 207 g mol^-1
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)

= 121.14 × 10^-24 cm³
Edge length, a = (121.14)^1/3 × 10^-8
= 4.948 × 10^-8 cm
or = 4.948 × 10^-10 m
or = 494.8 × 10^-12 m = 494.8 pm
Now, radius in fcc = \(\frac{a}{2 \sqrt{2}}\)
= \(\frac{494.8 \mathrm{pm}}{2 \times 1.414}\) = 174.96 pm

Question 6.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain. (C.B.S.E. 2011)
Answer:
Consider a unit cell of edge a cm
Volume of unit cell = a³ cm³

Mass of unit cell = No. of atoms in a unit cell × Mass of each atom = Z × m …… (ii)
Mass of an atom present in a unit cell

Knowing density (d), edge length (a), number of atoms per unit cell (Z) and Avogadro’s number (6.02 × 10²³), atomic mass can be calculated.

Question 7.
ZnO turns yellow on heating. Why? (C.B.S.E. 2016)
Answer:
When ZnO is heated, it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect. Due to the presence of free electrons in the interstitial sites the colour is yellow.

Question 8.
(a) What type of semiconductor is obtained when silicon is doped with boron?
Answer:
(a) When silicon is doped with boron having three valence electrons, the bonds formed create electron deficient sites called holes. Under the influence of applied electric field, one electron from neighbouring atom moves to fill the hole but creates another hole at its own place. Therefore, the electrical conductance is due to movement of positive holes. Therefore, this type of semi-conductor is called p-type semi-conductor.

(b) What type of magnetism is shown in the following alignment of magnetic moments?
↑    ↑    ↑    ↑   ↑   ↑    ↑
Answer:
Ferromagnetic.

(c) What type of point defect is produced when AgCI is doped with CdCl₂? (C.B.S.E. Delhi 2013)
Answer:
CdCl₂ on adding to AgCl introduces impurity defect. The addition of one Cd²⁺ ion will replace two Ag⁺ ions to maintain electrical neutrality. One of the positions of Ag⁺ will be occupied by Cd²⁺ ion and the other will be left as a hole. Thus, a hole is created similar to Schottky defect.

Question 9.
Account for the following:
(i) Schottky defects lower the density of related solids.
Answer:
In Schottky defect, there are holes due to missing cations and anions. Due to the presence of holes in solid, the density decreases.

(ii) Conductivity of silicon increases on doping it with phosphorus. (C.B.S.E. 2013)
Answer:
Pure silicon has a network lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus (s²p³) having five valence electrons, the impurity leads to excess of electrons after forming four covalent bonds like silicon. The extra electrons serve to conductivity and therefore, the conductivity of silicon doped with phosphorus increases.

Question 10.
Examine the given defective crystal

Answer the following questions:
(i) What type of stoichiometric defect is shown by the crystal?
Answer:
Schottky defect

(ii) How is the density of the crystal affected by this defect?
Answer:
Density of the crystal decreases.

(iii) What type of ionic substances show such defect? (C.B.S.E. Delhi 2014)
Answer:
Crystals having
(a) high coordination number and
(b) ions (cations and anions) of almost similar sizes. For example: KCl, KBr.

Question 11.
An element crystallises in a fcc lattice with edge length of 400 pm. The density of the element is 7 g cm^-3. How many atoms are present in 280 g of the element? (C.B.S.E. 2016)
Answer:
Edge length, a = 400 pm
= 400 × 10^-10 cm
Volume of unit cell = (400 × 10^-10)³
= 64 × 10^-24 cm^-3
Mass of element = 280 g
Density = 7 g cm^-3
Volume of 280 g of the element = \(\frac{280 \mathrm{~g}}{7 \mathrm{~g} \mathrm{~cm}^{-3}}\)
= 40 cm³
No. of unit cells in this volume
= \(\frac{40 \mathrm{~cm}^{3}}{64 \times 10^{-24} \mathrm{~cm}^{3}}\)
= 6.25 × 10²³ unit cells
Since the structure is fcc, number of atoms in a unit cell = 4
∴ No. of atoms in 280 g of element
= 4 × 6.25 × 10²³
= 2.5 × 10²⁴ atoms

Question 12.
An element crystallises in a fcc lattice with cell edge of 250 pm. Calculate the density of 300 g of this element containing 2 × 1024 atoms. (C.B.S.E. Delhi 2016)
Answer:
Length of edge, a = 250 pm = 250 × 10^-10 cm
Volume of unit cell = (250 × 10^-10)³
= 15.625 × 10^-24 cm³
Mass of unit cell = No. of atoms in unit cell × mass of each atom
Since the element has fcc arrangement, number of atoms per unit cell, Z = 4
Mass of an atom = \(\frac{300}{2 \times 10^{24}}\) = 1.50 × 10^-22 g

Question 13.
An element crystallises in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm^-3. Calculate the number of atoms in 108 g of the element. (C.B.S.E. Delhi 2019)
Answer:
Edge length of the unit cell = 300 pm
= 300 × 10^-10 cm
Volume of unit cell = (300 × 10^-10) ³
= 27 × 10^-24 cm³
Mass of element = 108 g
Volume of 108 g element = \(\frac{108 \mathrm{~g}}{10.8 \mathrm{~g} \mathrm{~cm}^{-3}}\)
= 10 cm³
No. of unit cells = \(\frac{10}{27 \times 10^{-24}}\)
= 3.7 × 10²³ unit cells
Since the structure is fcc, number of atoms per unit cell = 4
Number of atom is 108 g = 3.7 × 10²³ × 4
= 1.48 × 10²⁴.

Question 14.
(a) An element crystallises in bcc lattice with a cell edge of 3 × 10^-8 cm. The density of the element is 6.89 g cm^-3. Calculate the molar mass of the element. (N_A = 6.022 × 10²³ mol^-1)
Answer:
Density of element, d = 6.89 gcm^-3
Cell edge, a = 3 × 10^-8 cm
Z = 2 (bcc)

(b) What type of semiconductor is obtained when
(i) Ge is doped with In?
(ii) Si is doped with P? (C.B.S.E. 2019 Al)
Answer:
(i) p-type
(ii) n-type

Question 15.
(a) A compound is formed by two elements M and N. The element N forms ccp and M atoms occupy 1/3rd of tetrahedral voids. What is the formula of the compound?
(b) Which of the following lattices has the highest packing efficiency
(i) simple cubic
(ii) body centred cubic and
(iii) hexagonal close packed lattice?
(c) An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If the density is 2.7 × 10³ kg m^-3. What is the nature of the cubic unit cell?
Answer:
(a) Since N forms ccp arrangement, it will have 4 atoms in a unit cell.
Number of N atoms in unit cell = 4
For each atom, there are two tetrahedral voids so that there are 8 tetrahedral voids per unit cell.
No. of M atoms = \(\frac{1}{3}\) × 8 = \(\frac{8}{3}\)
Formula = M_8/3N₄
or = M₂ N₃.

(b) The packing efficiencies are:
simple cubic = 52.4%
body centred cubic = 68%
hexagonal close packed = 74%
∴ Hexagonal close packed lattice has highest packing efficiency.

(c) Edge length = 405 pm = 405 × 10^-12 m
Density of the cell = 2.7 × 10^-3 kg m^-3
Molar mass = 2.7 × 10^-2 kg mol^-1

Since the unit cell contains 4 atoms, it is cubic close packed structure, ccp.

Question 16.
How will you distinguish between the following pair of terms:
(i) Hexagonal close packing and cubic close packing
(ii) Crystal lattice and unit cell.
(iii) Tetrahedral void and octahedral void.
Answer:
(i)

(ii) The three dimensional arrangement of constituent particles of a substance (atoms, ions or molecules) is called crystal lattice.
The smallest repeating pattern in a crystal lattice which when repeated in three dimensions gives the crystal is called unit cell.

(iii) A void surrounded by four spheres is called a tetrahedral void while a void surrounded by six spheres is called an octahedral void.

Question 17.
(i) How many lattice points are there in one unit cell of each of the following lattice?
(a) face centred cubic
(b) face centred tetragonal
(c) body centred.
Answer:
(i) (a) In face centred cubic arrangement, number of lattice points are:
8 (at corners) + 6 (at face centres)
Lattice points per unit celt
= 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) = 4

(b) In face centred tetragonal, number of lattice points are:
= 8 (at corners) + 6 (at face centres) Lattice points per unit cell 1 1
= 8 × \(\frac{1}{8}\) + 6 × \(\frac{1}{2}\) = 4

(c) In body centred cubic arrangement number of lattice points are:
= 8 (at corners) + 1 (at body center) Lattice points per unit cell
= 8 × \(\frac{1}{8}\) + 1 = 2

(ii) Explain: (a) The basis of similarities and differences between metallic and ionic crystals.
(b) Ionic solids are hard and brittle.
Answer:
(a) Basis of similarities
1. Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these are between the oppositely charged ions while in metals, these are among the valence electrons and the kernels. That is why both ionic and metallic crystals have high melting points.
2. In both cases, the bond is non- directional.

Basis of differences
1. In ionic crystals, the ions are not free to move and therefore, they do not conduct electricity in the solid state. They conduct electricity in the molten state or in their aqueous solution. However, in metals, the valence electrons are free to move and hence they conduct electricity in the solid state.
2. Ionic bond in ionic crystals is strong due to electrostatic forces of attraction. However, metallic bond may be weak or strong depending upon the number of valence electrons and the size of the kernels.

(b) Ionic crystals are hard because there are strong electrostatic forces of attraction among the oppositely charged ions. They are brittle because the ionic bond is non-directional.

Question 18.
(a) If the radius of the octahedral void is r and radius of the atoms in close packing is R. Derive relation between r and R.
Answer:
A sphere of radius r filling in an octahedral void of spheres of radius R is shown in figure.

If the length of the unit cell is a cm, then
In right angled ∆ABC, AB = BC = a cm
The diagonal AC is:

Thus, for an atom to occupy an octahedral void, its radius must be 0.414 times the radius of the sphere.

(b) Copper crystallises into a fcc lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.29 g cm^-3.
Answer:
Density, d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
For fcc lattice, Z = 4
Atomic mass, M of copper = 63.5 g/mol^-1
a = 3.61 × 10^-8 cm
∴ p = \(\frac{4 \times 63.5}{\left(3.61 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}\)
= 8.96 g cm^-3
This value is close to measured value.

Question 19.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
The substances whose conductance lies between that of conductors (metals) and insulators are called semiconductors. They have conductivity values ranging from 10^-6 to 10⁴Ω^-1m^-1. Two main types of semiconductors are n-type and p-type.

(i) n-type semiconductors: These are the semiconductors in which the current is carried by the electrons in the normal way. For example, silicon and germanium belong to group 14 of the periodic table and have four valence electrons each. In their crystals, each atom forms four covalent bonds with its neighbours.

When it is doped with an element of group 15 such as P or As, which contains five valence electrons, they occupy some of the lattice sites in silicon or germanium crystal. Four of the five valence electrons are used in the formation of four covalent bonds with the four neighbouring Si atoms and the fifth electron is extra and becomes delocalised.

This delocalised electron will be able to conduct electricity and such type of conduction is called n-type conduction. Hence silicon or germanium doped with extra electron of impurity are called n-type semiconductors.

(ii) p-type semiconductors: These are the semiconductors in which current is carried by the movement of positive holes. For example, when Ge or Si are doped with electron deficient atoms such as Ga or In (of Group 13) containing three valence electrons, the atoms of Ga or In will replace Ge atoms.

Each In or Ga atom will use its three electrons for forming three covalent bonds with neighbouring Ge atoms and the place for fourth bond will remain missing and is called electron vacancy or hole. Such holes can move through the crystals like a positive charge giving rise to conductivity. This type of conduction is called p-type conduction. Hence, Si or Ge doped with electron deficient atoms as impurity are called p-type semiconductors.

Question 20.
Explain the following with suitable examples:
(a) Ferromagnetism
(b) Paramagnetism
(c) Ferrimagnetism
(d) Anti-ferromagnetism
(e) 12-16 and 13-15 group compounds.
Answer:
(a) Ferromagnetism: When there is spontaneous alignment of magnetic moments of domains in the same direction, we get ferromagnetism. These have strong magnetic effects and ordering of domains persists even when magnetic field is removed and – ferromagnetic substance becomes permanent magnet.

(b) Paramagnetism: The substances which have permanent magnetic dipoles and are attracted by the magnetic field are called paramagnetic substances. This property of attraction by the magnetic field is known as paramagnetism. The paramagnetic substance are atoms, molecules or ions having unpaired electrons. For example, O₂, Cu²⁺, Fe²⁺, etc. They lose their magnetism in the absence of magnetic field.

(c) Ferrimagnetism: When the magnetic moments of the domains are aligned in parallel or anti-parallel directions in unequal numbers resulting in net magnetic moment, we get ferrimagnetism.

(d) Anti-ferromagnetism: It arises when the alignment of magnetic moments is in a compensatory way so that the resultant magnetic moment is zero (e.g., MnO).

(e) The solid binary compounds prepared by combining elements of group 12 and 16 are called 12-16 compounds. For example CdS, ZnS, etc. The compounds prepared by combining elements of group 13 and 15 are called 13-15 compounds. For example, AlP, GaAs, etc. These compounds are used as semiconductors.

Question 21.
Give the points of differences between Schottky defect and Frenkel defect ?
OR
Explain:
(i) Why is Frenkel defect not found in pure alkali metal halides ?
(ii) Zinc oxide is white but it turns yellow on heating.
(iii) CaCl₂ will introduce Schottky defect when added to AgCl crystal.
Answer:
The important differences between Schottky and Frenkel defects are given ahead:

OR
(i) Frenkel defect is not found in pure alkali metal halides because the ions cannot get into the interstitial sites due to their larger size.

(ii) When ZnO is heated it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect. Due to the presence of free electrons in the interstitial sites the colour is yellow.

(iii) CaCl₂ on adding to AgCl introduces impurity defect. The addition of one Ca²⁺ ion will replace two Ag⁺ ions to maintain electrical conductivity. One of the position of Ag⁺ will be occupied by Ca²⁺ ion and other will be left as a hole. Thus, a hole is created similar to Schottky defect.

Question 22.
What do you understand by a space lattice and a unit cell?
Answer:
(a) Space lattice is a regular arrangement of the constituent particles (atoms, ions or molecules) of a crystalline solid in three dimentional space.

The positions which are occupied by atoms, ions or molecules in the crystal lattice are called lattice points or lattice sites. A two dimensional and a three dimensional space lattice is shown in Fig. 1 (a) and 1 (b) respectively.


Fig. 1. (a) Two dimensional and
(b) Three dimensional space lattice.

A unit cell is the smallest repeating unit in space lattice which when repeated over and over again results in the crystal of the given substance. The unit cell gives the shape of the entire crystal. The crystal may be considered to consist of an infinite number of unit cells. The unit cell in the above crystal lattice [Fig. 1 (b)] is shown in The complete crystal lattice can be obtained by extending the unit cell in all the three directions.

Question 23.
What are tetrahedral and octahedral holes (voids) in close packed stacks of spheres? Explain.
Answer:
Tetrahedral and octahedral holes. The close packed arrangement of spheres has two types of holes or voids:
Tetrahedral holes: Tetrahedral hole Is produced when one sphere rests upon three other touching spheres. The space which is left between these touching spheres is called tetrahedral hole or tetrahedral site. The tetrahedral hole is shown in Fig. (a). It may be noted that in close packed arrangement (ccp or hcp), there are twice as many tetrahedral holes as there are spheres.

Octahedral hole.: This type of hole is formed at the centre of six spheres which form a regular octahedral. This is shown in Fig. (b). From the figure, it is clear that the octahedral hole is formed by two sets of equilateral triangles which point in opposite directions. It may be noted that there are same number of octahedral holes as the number of spheres.

Fig. (a) Tetrahedral hole (b) Octahedral hole.

Question 24.
(a) Why does table salt, NaCl, some times appear yellow in colour?
(b) Why is FeO(s) not formed in stoichiometric composition?
(c) Why does white ZnO (s) becomes yellow upon heating?
Answer:
(a) The yellow colour of sodium chloride crystals is due to metal excess defect. In this defect, the unpaired electrons get trapped in anion vacancies. These sites are called F-centres. The yellow colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.

(b) In the crystals of FeO, some of the Fe²⁺ cations are replaced by Fe³⁺ ions. To balance the charge, three Fe²⁺ ions are replaced by two Fe³⁺ ions to make up for the loss of positive charge. As a result, there would be less amount of metal as compared to stoichiometric proportion.

(c) When ZnO is heated it loses oxygen as:

The Zn²⁺ ions are entrapped in the interstitial sites and electrons are entrapped in the neighbouring interstitial sites to maintain electrical neutrality. This results in metal excess defect and F-centres are created. Due to the presence of electrons in the interstitial voids, the colour is yellow.

Question 25.
(a) Explain why does conductivity of germanium crystals increase on doping with gallium.
(b) In a compound, nitrogen atoms (N) make cubic close packed lattice and metal atoms (M) occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by M and N?
(c) Under which situations can an amorphous substance change to crystalline form?
Answer:
(a) On doping germanium with gallium, some of the positions of lattice of germanium are occupied by gallium. Gallium atom has only three valence electrons. Therefore, fourth valency of nearby germanium atom is not satisfied and this site remains vacant. This place is deficient of electrons and is called electron hole or electron vacancy.

Electron from neighbouring atom moves to fill the gap, thereby creating a hole in its original position. Under the influence of electric field, electrons move towards positively charged plates through these holes and conduct electricity. The holes appear to move towards negatively charged plates. The movement of electrons (or electron holes) results in increase in conductivity of germanium.

(b) N atoms make up ccp arrangement and there are two tetrahedral sites per atom of N.
No. of N atoms per unit cell = 4
No. of tetrahedral sites = 4 × 2 = 8
No. of sites occupied by M = \(\frac{1}{3}\) × 8 = \(\frac{8}{3}\)
∴ No. of M atoms per unit cell = \(\frac{8}{3}\)
Formula M_8/3N₄ or M₈N₁₂ or M₂N₃

(c) On heating, amorphous solids become crystalline at some temperature. For example, some glass objects from ancient civilisations are found to become milky in appearance because of some crystallisation.

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 2 Electrostatic Potential and Capacitance. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 2 Important Extra Questions Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance Important Extra Questions Very Short Answer Type

Question 1.
Express dielectric constant in terms of the capacitance of a capacitor.
Answer:
It is given by the expression K = \(\frac{c}{c_{0}}\) where C is the capacitance of the capacitor with dielectric and C₀ is the capacitance without the dielectric.

Question 2.
On what factors does the capacitance of a parallel plate capacitor depend?
Answer:

1. Area of plates,
2. The separation between the plates and
3. Nature of dielectric medium between the plates.

Question 3.
What is the ratio of electric field intensities at any two points between the plates of a capacitor?
Answer:
The ratio is one, as the electric field is the same at all points between the plates of a capacitor.

Question 4.
Write a relation between electric displacement vector D and electric field E.
Answer:
\(\vec{D}\) = ε₀ \(\vec{E}\) + \(\vec{P}\)

Question 5.
Write the relation between dielectric constant (K) and electric susceptibility χ_e
Answer:
K = 1 + χ_e

Question 6.
A hollow metal sphere c radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the center of the sphere? (CDSE AI 2011)
Answer:
10 V

Question 7.
What is the geometrical shape of equipotential surfaces due to a single isolated charge? (CBSE Delhi 2013)
Answer:
Concentric circles.

Question 8.
Draw the equipotential surfaces due to an isolated point charge. (CBSE Delhi 2019)
Answer:
These areas are shown.

Question 9.
‘For any charge configuration, equipotential surface through a point is normal to the electric field’. Justify. (CBSE Delhi 2014)
Answer:
This is because work done in moving a charge on an equipotential surface is zero. This is possible only if the equipotential surface is perpendicular to the electric field.

Question 10.
The given graph shows the variation of charge ‘q’ versus potential difference ‘V for two capacitors C₁ and C₂. Both the capacitors have the same plate separation but the plate area of C₂ is greater than that of C_y Which line (A or B) corresponds to C₁ and why? (CBSEAI 2014C)

Answer:
Since C = ε₀ A/d, since the area for C₂ is more, therefore capacitance of C₂ is more. From the graph greater the slope greater is than the capacitance, therefore, graph A belongs to capacitor C₂. While graph B belongs to capacitance C_v

Question 11.
Write a relation for polarisation P of dielectric material in the presence of an external electric field E . (CBSE AI 2015)
Answer:
P = χ_e ε₀ \(\vec{E}\).

Question 12.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE AI 2015C)
Answer:
So that no net force acts on the charge on the equipotential surface and it remains stationary.

Question 13.
Why does a given capacitor store more charge at a given potential difference when a dielectric is filled in between the plates?
Answer:
Because the capacitance of the capacitor increases on filling the dielectric medium in between the plates.

Question 14.
Why is the electrostatic potential inside a charged conducting shell constant throughout the volume of the conductor? (CBSE AI 2019)
Answer:
We know that E = – \(\frac{d V}{d r}\)
Inside a conductor, the electric field is zero and no work is done in moving a charge inside a conductor. Therefore, V is constant.

Question 15.
Does the charge given to a metallic sphere depend on whether it is hollow or solid? (CBSE Delhi 2017)
Answer:
If the capacitance of the two spheres, solid and hollow, is the same, then they will hold the same quantity of charge.

Question 16.
Is the electric potential necessarily zero at a place where the electric field is zero?
Answer:
No, it is not necessary. The electric field inside a hollow metallic conductor is zero but the electric potential is not zero.

Question 17.
In a parallel plate capacitor, the potential difference of 10² V is maintained between the plates. What will be the electric field at points A and B as shown in the figure below?
Answer:
The electric field between the plates of a capacitor is uniform; therefore the electric field at points A and B will be the same.

Question 18.
If the plates of a charged capacitor are suddenly connected to each other by a wire, what will happen?
Answer:
The capacitor is discharged immediately.

Question 19.
Can you place a parallel plate capacitor of one farad capacity in your house?
Answer:
Ordinarily, it is not possible because the surface area of such a capacitor will be extra-large.

Question 20.
Is there any conductor which can be given almost unlimited charge?
Answer:
Yes, the earth.

Question 21.
What would be the work done if a point charge + q is taken from a point A to a point B on the circumference of a circle drawn with another point charge + q at the center?

Answer:
Zero.

Question 22.
Sketch a graph to show how a charge Q, acquired by a capacitor of capacitance, C, varies with the increase in the potential difference between the plates.
Answer:

Question 23.
In a parallel plate capacitor, the capacitance increases from 4 μF to 80 μF, introducing a dielectric medium between the plates. What is the dielectric constant of the medium?
Answer:
The dielectric constant is given by
K = \(\frac{80}{4}\) = 20

Electrostatic Potential and Capacitance Important Extra Questions Short Answer Type

Question 1.
Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q. (CBSE Delhi 2012)
Answer:
The plot is as shown.

Question 2.
Two identical capacitors of 10 pF each are connected in turn (i) in series and (ii) in parallel across a 20 V battery. Calculate the potential difference across each capacitor in the first case and the charge acquired by each capacitor in the second case. (CBSE AI 2019)
Answer:
(i) Since the two capacitors have the same capacitance, therefore, the potential will be divided amongst them. Hence V = 10 V each
(ii) Since the capacitors are connected in parallel, therefore, potential difference = 20 V
Hence charge Q = CV = 10 × 20 = 200 pC

Question 3.
A point charge ‘q’ is placed at O as shown in the figure. Is V_A – V_B positive, negative, or zero, if ‘q’ is an (i) positive, (ii) negative charge? (CBSE Delhi 2011, 2016)

Answer:
If V_A – V_B = \(\frac{q}{4 \pi \varepsilon_{0}}\left(\frac{1}{\mathrm{OA}}-\frac{1}{\mathrm{OB}}\right)\)

As OA < OB
∴ If q is positive then V_A– V_B is positive and
if q is negative V_A – V_B is also negative.

Question 4.
The graph shows the variation of voltage V across the plates of two capacitors A and B versus charge Q stored on them. Which of the two capacitors has higher capacitance? Give a reason for your answer.

Answer:
Capacitor A has higher capacitance. We know that capacitance C = Q/V.

For capacitor A
\(c_{A}=\frac{Q}{V_{A}}\)

For capacitor B
\(c_{B}=\frac{Q}{V_{B}}\)

As V_B > V_A
∴ C_B < C_A
Thus capacitance of A is higher.

Question 5.
A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field E as shown in the figure,

(i) Calculate the potential difference between A and C
Answer:
(i) dV = – E dr = – E (6 – 2) = – 4E

(ii) At which point (of the two) is the electric potential more and why? (CBSE AI 2012)
Answer:
Electric potential is more at point C as dV = – Edr, i.e. the electric potential decreases in the direction of the electric field.

Question 6.
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor. (CBSE AI 2013)
Answer:
Given t = d/2, C = ?
We know that when a dielectric of thickness ‘t’ is inserted between the plates of a capacitor, its capacitance is given by
C = \(\frac{\varepsilon_{0} A}{d-t+\frac{t}{K}}\)

Hence we have
C = \(\frac{\varepsilon_{0} A}{d-\frac{d}{2}+\frac{d}{2 K}}=\frac{2 K \varepsilon_{0} A}{d(1+K)}\)

Question 7.
Two-point charges q and -2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 8.
Four-point charges Q, q, Q., and q are placed at the corners of a square of side ‘a’ as shown in the figure.

Find the potential energy of this system. (CBSEAI, Delhi 2018)
Answer:
The potential energy of the system
U = \(\frac{1}{4 \pi \varepsilon_{0}}\left(4 \frac{q Q}{a}+\frac{q^{2}}{a \sqrt{2}}+\frac{Q^{2}}{a \sqrt{2}}\right)\)

U = \(\frac{1}{4 \pi \varepsilon_{0} a}\left(4 q Q+\frac{q^{2}}{\sqrt{2}}+\frac{Q^{2}}{\sqrt{2}}\right)\)

Question 9.
Three-point charges q, -4q, and 2q are placed at the vertices of an equilateral triangle ABC of side T as shown in the figure. (CBSE Delhi 2018)

Find out the amount of the work done to separate the charges at infinite distance. (CBSE AI, Delhi 2018)
Answer:
Net potential energy of the system
= \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{l}[-4+2-8]=\frac{5 q^{2}}{2 \pi \varepsilon_{0} l}\)

Electrostatic Potential and Capacitance Important Extra Questions Long Answer Type

Question 1.
Two-point charges 2 μC and —2 μC are placed at points A and B 6 cm apart.
(a) Draw the equipotential surfaces of the system.
Answer:
The diagram is as shown.

(b) Why do the equipotential surfaces get closer to each other near the point charges? (CBSEAI2O11C)
Answer:
We know that E = – dV/dr
Therefore, dr =- dV/E
Since near the charge, electric field E is large, dr will be less.

Question 2.
(a) Obtain the expressions for the resultant capacitance when the three capacitors C₁, C_2, and C₃ are connected (i) in parallel and then (ii) in series.
Answer:
(i) Parallel combination of three capacitors.
Let three capacitors of capacitances C₁, C₂, and C₃ be connected in parallel, and potential difference V be applied across A and B. If q be total charge flowing in the circuit and q₁ q₂ and q₃ be charged flowing across C₁, C_2, and C₃ respectively, then
q = q₁+ q₂ + q₃
or q = C₁v + C₂V + C₃V …(i)

If C_P is the capacitance of the arrangement in parallel, then
q = C_PV

So equation (i) becomes
C_PV = C₁V + C₂V + C₃V
Or
C_P = C₁ + C₂ + C₃

(ii) Series combination of three capacitors Let three capacitors C₁, C₂, and C₃ be connected in series. Let q charge be flowing through the circuit.
If V₁, V₂, and V₃ be potential differences across the plates of the capacitor and V be the potential difference across the series combination, then

V = V₁ + V₂ + V₃
Or
V = \(\frac{q}{C_{1}}+\frac{q}{C_{2}}+\frac{q}{C_{3}}\) … (i)

If Cs is the capacitance of series combination, then V = \(\frac{q}{\mathrm{C}_{\mathrm{s}}}\).

So the equation (i) becomes
\(\frac{q}{\mathrm{C}_{\mathrm{s}}}=\frac{q}{\mathrm{C}_{1}}+\frac{q}{\mathrm{C}_{2}}+\frac{q}{\mathrm{C}_{3}}\)
Or
\(\frac{1}{\mathrm{C}_{\mathrm{s}}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\frac{1}{\mathrm{C}_{3}}\)

(b) In the circuit shown in the figure, the charge on the capacitor of 4 μF is 16 μC. Calculate the energy stored in the capacitor of 12 μF capacitance. (CBSE 2019C)

Answer:
Charge q across 4 μF Capacitor is 10 μc Potential difference across the capacitor of capacitance 4 μF will be
V= \(\frac{q}{C}=\frac{16 \mu C}{4 \mu F}=\frac{16 \times 10^{-6} \mathrm{C}}{4 \times 10^{-6} \mathrm{~F}}\)=4V

∴ Potential across 12 μF Capacitors
= 12V – 4V = 8V

Energy stored in the capacitors of capacitance C = 12 μF

U = \(\frac{1}{2}\) CV² = \(\frac{1}{2}\) × 12 × 10^-6 × 8² joule
= 384 × 10^-6 J = 384 μJ

Question 3.
Two identical plane metallic surfaces A and B are kept parallel to each other in the air, separated by a distance of 1 cm as shown in the figure.

A is given a positive potential of 10 V and the outer surface of B is earthed.
(i) What is the magnitude and direction of the uniform electric field between Y and Z?
Answer:
The electric field between the plates is
E = \(\frac{V}{d}\) = 10³ V m^-1
directed from plate A at the higher potential to plate B at a lower potential, i.e. from Y to Z

(ii) What is the work done in moving a charge of 20 µC from X to Y?
Answer:
Since X and Y are on the same plate A, which is an equipotential surface, work done in moving a charge of 20 µC from X to Y on the equipotential surface is zero.

Question 4.
(i) Draw the equipotential surfaces corresponding to a uniform electric field in the z-direction.
Answer:
The equipotential surfaces are as shown.

Equipotential surfaces

(ii) Derive an expression for the electric potential at any point along the axial line of an electric dipole. (CBSE Delhi 2019)
Answer:
Consider an electric dipole of length 2a and having charges +q and -q. Let us find the potential on the axial line at point P at a distance OP = x from the center of the dipole.

Now potential at point P is

Question 5.
A network of four capacitors, each of capacitance 15 µF, is connected across a battery of 100 V, as shown in the figure. Find the net capacitance and the charge on the capacitor C₄. (CBSE Al 2012C)

Answer:
Capacitors C1 C2 and C3 are in series, therefore their equivalent capacitance is
\(\frac{1}{C_{\mathrm{s}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}=\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{3}{15}\)

Hence C_S = 5 µF
Now C_S and C₄ are in parallel, hence
C_P = C_S + C₄ = 5 + 15 = 20 µF

Now C4 is connected to 100 V, therefore charge on it is
Q = CV=15 × 10^-6 × 100=15 × 10^-4 C

Question 6.
A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor. (CBSE AI 2014)
Answer:
U_i = \(\frac{1}{2}\) CV²

When the capacitors are connected then the energy stored is
U_F = \(\frac{1}{2}\left(C_{1}+C_{2}\right)\left(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\right)^{2}\)

Since C₁ = C₂ = C, and V₂ = 0, we have
Uf = \(\frac{1}{2}(C+C)\left(\frac{C V}{C+C}\right)^{2}=\frac{1}{2}(2 C) \times \frac{V^{2}}{4}=\frac{1}{4} C V^{2}\)

Hence we have
\(\frac{U_{f}}{U_{i}}=\frac{1}{2}\)

Question 7.
An isolated air capacitor of capacitance C₀ is charged to a potential V₀. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how to do the following change, when the battery remains connected
(i) capacitance,
Answer:
When the battery remains connected, the potential on the capacitor does not change.
The capacitance of the capacitor becomes K times the original value, i.e. C = K C₀.

(ii) charge
Answer:
Now new charge is Q = CV = K C₀ V = K Q₀.

(iii) the field between the plates
Answer:
The field between the plates becomes
E = \(\frac{V}{d}=\frac{V_{0}}{d}\) = E₀, i.e. no change.

(iv) energy stored by the capacitor?
Answer:
The energy stored becomes
U = \(\frac{1}{2} C V^{2}=\frac{1}{2} K C_{0} V^{2}\) = KU₀.

Question 8.
An isolated air capacitor of capacitance C0 is charged to a potential V0. Now if a dielectric slab of dielectric constant K is inserted between its plates, completely filling the space between the plates, then how to do the following change, when the battery is disconnected
(i) charge
Answer:
If battery is disconnected then charge remains same, Q = Q₀

(ii) electric field between the plates,
Answer:
V = \(\frac{V_{0}}{K}\) ,
∴E = \(\frac{E_{0}}{K}\) ,

(iii) capacitance
C = KCO [∵ C = \(\frac{Q_{0}}{V}=\frac{Q_{0}}{V_{0} / K}=\frac{K Q_{0}}{V_{0}}\)]

(iv) energy stored by the capacitor
U = \(\frac{U_{0}}{K}\left[U=\frac{1}{2} C V^{2}=\frac{1}{2}\left(K C_{0}\right)\left(\frac{V_{0}}{K}\right)^{2}\right]\)

Question 9.
The figure shows two identical capacitors, C₁ and C₂, each of 1 µF capacitance connected to a battery of 6 V. Initially switch ‘S’ is closed. After some time ‘S’ is left open and dielectric slabs of dielectric constant K = 3 are inserted to fill completely the space between the plates of the two capacitors. How will (i) the charge and (II) potential difference between the plates of the capacitors be affected after the slabs are inserted? (CBSE Delhi 2011)

Answer:
When switch S is opened then capacitor C₁ remains connected to the battery white capacitor C₂ is disconnected. Thus for the two capacitors, we have

Question 10.
A particle, having a charge +5 µC, is initially at rest at the point x = 30 cm on the x axis. The particle begins to move due to the presence of a charge Q that is kept fixed at the origin. Find the kinetic energy of the particle at the instant it has moved 15 cm from its initial position if (i) Q = +15 µC and (ii) Q = -15 µC (CBSE Sample Paper 2018-19)
Answer:
From energy conservation, U_i + K_i = U_f + K_f
kQq/r_i + 0 = kQq/r_f + K_f
K_f = kQq (1/r_i — 1/r_f)

When Q is +15 µC, q will move 15 cm away from it. Hence r_f = 45 cm
K_f = 9 × 10⁹ × 15 × 10^-6 × 5 × 10^-6 [1/(30 × 10^-6) – 1/(45 × 10^-2)] = 0.75 J

When Q is -15 µC, q will move 15 cm towards it. Hence r_f = 15 cm K_f = 9 × 10⁹ × (-15 × 10^-6) × 5 × 10⁶ [1/(30 × 10^-2) – 1/(15 × 10^-2)] = 2.25 J

Question 11.
Two metal spheres, one of radius R and the other of radius 2R, both have same surface charge density σ. They are brought in contact and separated. What will be new surface charge densities on them? (NCERT Exemplar)
Answer:
Let the two spheres have charges Q₁ and Q₂ respectively. Since σ₁ = σ₂, before contact, we have
\(\frac{Q_{1}}{4 \pi R^{2}}=\frac{Q_{2}}{4 \pi(2 R)^{2}}\)
Or
Q₂ = 4 Q₁

After contact
Let q1 and q2 be the charges on them, then
q₁ + q₂ = Q₁ + Q₂= Q₁ + 4Q₁ = 5Q₁ = 5(σ × 4πr²)

The two will exchange charge till their potentials are equal, therefore we have
\(\frac{q_{1}}{R}=\frac{q_{2}}{2 R}\)
Or
q₂ = 2 q₁

Therefore 3q₁ = 5(σ × 4πr²)
Or
q₁ = \(\frac{5}{3}\)(σ x 4πr²) and q₂ = 2q₁ = \(\frac{10}{3}\)5(σ x 4πr²)

Therefore
σ₁ = \(\frac{q_{1}}{4 \pi R^{2}}=\frac{5}{3}\left(\frac{\sigma \times 4 \pi R^{2}}{4 \pi R^{2}}\right)=\frac{5 \sigma}{3}\)
And
σ₂ = \(\frac{q_{2}}{4 \pi(2 R)^{2}}=\frac{10}{3}\left(\frac{\sigma \times 4 \pi R^{2}}{4 \pi(2 R)^{2}}\right)=\frac{5 \sigma}{6}\)

Question 12.
(a) Derive an expression for the capacitance of a parallel plate capacitor when the space between the plates is partially filled with a dielectric medium of dielectric constant ‘K’.
Answer:
Consider a parallel plate capacitor having each plate of area A and separated by a distance d. When there is a vacuum between the two plates, the capacitance of the parallel plate capacitor is given by
C₀ = ε₀ A/d.

Suppose that when the capacitor is connected to a battery, the electric field of strength E₀ is produced between the two plates of the capacitor. Further, suppose that when a dielectric slab of thickness t (t < d) is introduced between the two plates of the capacitor as shown in the figure, the electric field reduces to E due to the polarisation of the dielectric. Therefore between the two plates of the capacitor, over a distance of t, the strength of the electric field is E, and over the remaining distance (d – f)

the strength is E₀. If V is the potential between the plates of the capacitor, then

V = E_t + E₀(d – t)
Since E = E₀/K where K is the dielectric constant, the above equation becomes

V = \(\frac{E_{0}}{K} t+E_{0}(d-t)=E_{0}\left(d-t+\frac{t}{K}\right)\)

The electric field between the plates of the capacitor is given by
E₀ = σ / ε₀ = Q/ A ε₀

Hence the potential between the two plates becomes.
V = \(E_{0}\left(d-t+\frac{t}{K}\right)=\frac{Q}{\varepsilon_{0} A}\left(d-t+\frac{t}{K}\right)\)

Hence the capacitance of the parallel plate capacitor is given

∴ clearly, C > C₀. Therefore, capacitance increases in the presence of a dielectric medium.

(b) Explain why the capacitance decreases when the dielectric medium is removed from between the plates.
Answer:
On removing the dielectric, the capacitance will decrease.

Question 13.
(a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor.
(b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop a b c d a. (CBSE Delhi 2014)

Answer:
(a) Suppose the capacitor is charged fully, its final charge is Q. and a final potential difference is V. These are related as Q = CV

Let q and V be the charge and potential difference, respectively.

At an intermediate stage during charging process q = CV. At this stage the small work done dW to transfer an additional charge dq is

dW = Vdq = \(\frac{q d q}{C}\)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by
W = \(\int_{0}^{w} d W=\int_{0}^{Q} \frac{q d q}{C}=\frac{1}{C} \int_{0}^{Q} q d q\)
Or
W = \(\frac{1}{C}\left|\frac{q^{2}}{2}\right|_{0}^{Q}=\frac{Q^{2}}{2 C}\)

This work is stored in the capacitor in the form of its electric potential energy. Hence,
U = \(\frac{Q^{2}}{2 C}\) … (1)

substituting Q = CV in equation (1) we have
U = \(\frac{1}{2}\) CV²

Energy density = Energy stored per unit volume

(b) Since the surface is an equipotential surface, work done is zero.

Question 14.
(a) Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
Answer:
Suppose Q. is the charge on the capacitor, and c is the uniform surface charge density on each plate as shown in the figure. Therefore by Gauss’s theorem, the electric field between the plates of the capacitor (neglecting fringing of electric field at the edges) is given by

E = \(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\) …. (1)

The field is uniform, and the distance between the plates is d, so the potential difference between the two plates is
V = Ed = \(\frac{1}{\varepsilon_{0}} \frac{Q d}{A}\) ….. (2)

Therefore by the definition of capacitance we have
C = \(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\) …. (3)

This gives the capacitance of a parallel plate capacitor with a vacuum between plates.

(b) Two charged spherical conductors of radii R₁ and R₂ when connected by a conducting wire acquire charges q₁ and q₂ respectively. Find the ratio of their surface charge densities in terms of their radii. (CBSE Delhi 2014)
Answer:

The ratio of the surface charge densities is given by
\(\frac{\sigma_{1}}{\sigma_{2}}=\frac{q_{1}}{4 \pi R_{1}^{2}} \times \frac{4 \pi R_{2}^{2}}{q_{2}}\) ….(4)

Now when they are connected with a wire, their potentiaLs wilt be same: therefore, from the expression.
V = \(\frac{Q}{c}=\frac{Q}{R}\)

∵ C = 4πε₀R for a spherical body
∴\(\frac{q_{1}}{q_{2}}=\frac{R_{1}}{R_{2}}\)

Substituting in equation (4) we have
\(\frac{\sigma_{1}}{\sigma_{2}}=\frac{R_{1}}{R_{1}^{2}} \times \frac{R_{2}^{2}}{R_{2}}=\frac{R_{2}}{R_{1}}\)

Question 15.
(a) Describe briefly the process of transferring the charge between the two plates of a parallel plate capacitor when connected to a battery. Derive an expression for the energy stored in a capacitor.
Answer:
The electrons are transferred to the positive terminal of the battery from the metallic plate connected to the positive terminal, leaving behind a positive charge on it. Similarly, the electrons move on to the second plate from the negative terminal, hence it gets negatively charged. This process continues till the potential difference between the two plates becomes equal to the potential of the battery.

Suppose the capacitor is charged fully; its final charge is Q and the final potential difference is V.
These are related as Q = CV

Let q and V be the charge and potential difference respectively, after some time during the charging of the capacitor, then q = CV. At this time, the small work done dW required to transfer an additional charge dq is given by

dW = Vdq = \(\frac{q d q}{C}\)

The total work W needed to increase the capacitor’s charge q from zero to its final value Q is given by

This work is stored in the capacitor in the form of its electric potential energy. Hence,
U = \(\frac{Q^{2}}{2 C}\) …(1)

Substituting Q= CVin equation (1) we have
U = \(\frac{1}{2}\) CV²

(b) A parallel plate capacitor is charged by a battery to a potential difference V. It is disconnected from the battery and then connected to another uncharged capacitor of the same capacitance. Calculate the ratio of the energy stored in the combination to the initial energy on the single capacitor. (CBSE Delhi 2019)
Answer:
The charge stored on the capacitor q = CV, when it is connected to the uncharged capacitor of same capacitance, sharing of charge, takes place between the two capacitors till the potential of both the capacitors becomes V/2.

Energy stored on the combination (U₂)
= \(\frac{1}{2} C\left(\frac{V}{2}\right)^{2}+\frac{1}{2} C\left(\frac{V}{2}\right)^{2}=C\left(\frac{V}{2}\right)^{2}=\frac{C V^{2}}{4}\)

Energy stored on single capacitor before connecting
U₁ = –\(\frac{1}{2}\) CV²

Ratio of energy stored in the combination to that in the single capacitor.
\(\frac{U_{2}}{U_{1}}=\frac{C V^{2} / 4}{C V^{2} / 2}=\frac{1}{2}\)

Question 16.
(a) Obtain the expression for the potential due to an electric dipole of dipole moment p at a point ‘x’ on the axial line.
(b) Two identical capacitors of plate dimensions l × b and plate separation d have dielectric slabs filled In between the space of the plates as shown In the figures.

Obtain the relation between the dielectric constants K, K_1, and K₂. (CBSE Al 2013C)
Answer:
(a) Consider an electric dipole of length 2a and having charges + q and — q. Let us find the potential on the axial Une at point P at a distance OP = x from the center of the dipole.

Now potential at point P is

(b) When there is no dieLectnc then
C = \(\frac{\varepsilon_{0} l b}{d}\)

For the first capacitor
C’ = \(\frac{K \varepsilon_{0} l b}{d}\) = KC

The second case is a case of two capacitors connected in paralleL, therefore
C₁ = \(\frac{K_{1} \varepsilon_{0} l b}{2 d}\) and
C₂ = \(\frac{K_{2} \varepsilon_{0} l b}{2 d}\)

These two are connected in parallel, therefore we have
C” = C₁ + C₂ = \(\frac{K_{1} \varepsilon_{0} l b}{2 d}\) + \(\frac{K_{2} \varepsilon_{0} l b}{2 d}\)
= C\(\left(\frac{K_{1}+K_{2}}{2}\right)\)

If the capacitance in each case be same, then C’ = C”

Hence K= \(\left(\frac{K_{1}+K_{2}}{2}\right)\)

Question 17.
(a) Explain using suitable diagrams the difference in the behavior of a
(i) conductor:
Answer:
The diagram is as shown.

When a conductor is placed in an external electric field, the free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor. This happens until in the static situation, the two fields cancel each other and the net electrostatic field in the conductor is zero.

(ii) dielectric In the presence of the external electric field. Define the term polarisation of a dielectric and write its relation with susceptibility.
Answer:

In a dielectric, this free movement of charges is not possible. It turns out that the external field induces dipole moment by stretching or re-orienting molecules of the dielectric. The collective effect of all the molecular dipole moments is that the net charges on the surface of the dielectric produce a field that opposes the external field. However, the opposing field so induced does not exactly cancel the external field. It only reduces it. The extent of the effect depends on the nature of the dielectric.

Dielectric polarization is defined as the dipole moment per unit volume of a dielectric. It is related to susceptibility as P = χ_eε₀\(\vec{E}\)

(b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge is placed at its center C and another charge +2Q. outside the shell, at a distance r from the center as shown in the figure. Find (i) the force on the charge at the center of the shell and at the point A and

Answer:
(i) F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{2 R^{2}}\) and
F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(Q+Q / 2) \times 2 Q}{r^{2}}\)

(ii) the electric flux through the shell. (CBSE Delhi 2015)
Answer:
Flux = \(\frac{Q}{2 \varepsilon_{0}}\)

Question 18.
(a) Define the SI unit of capacitance.
Answer:
The capacity of a capacitor is said to be one farad when a charge of 1 coulomb is required to raise the potential difference by 1 volt.

(b) Obtain the expression for the capacitance of a parallel plate capacitor.
Answer:
Let Q. be the charge on the capacitor, and o be the uniform surface charge density on each plate as shown in the figure.

d = distance between plates of the capacitor.

Therefore by Gauss’ theorem, the electric field between the plates of the capacitor is given by
E = \(\frac{\sigma}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}\)

The field is uniform, so the potential difference between the two plates is
V = Ed = \(\frac{1}{\varepsilon_{0}} \frac{Q d}{A}\)

Therefore, by the definition of capacitance we have
C = \(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\)

This gives the capacitance of a parallel plate capacitor.

(c) Derive the expression of the effective capacitance of a series combination of n capacitance (CBSE Delhi 2016C)
Answer:
Capacitors in series. Capacitors are said to be connected in series if the second plate of one capacitor is connected to the first plate of the next and so on as shown in the figure.

Resultant capacitance will be
\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\ldots \ldots \ldots \ldots=+\frac{1}{C_{n}}\)

Question 19.
A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R >> r), such that their surface charge densities are equal. Derive the expression for the potential at the common center.
Or
Three concentric metallic shells A, B, and C of radii a, b, and c (a <b < C) have surface charge densities +a, -a, and + o respectively as shown. Obtain the expressions for the potential of three shells A, B, and C. If shells A and C are at the same potential, obtain the relation between a, b and c. (CBSE Al 2019)

Answer:
Let the charges on the spheres be q, and q2 such that
Q=q₁ + q₂
= 4πσ(r² + R²)
Or
σ = \(\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)}\)

Now potential at the common centre V=V₁ + V₂
V = \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{1}}{r}+\frac{q_{2}}{R}\right)\)
= \(\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{4 \pi r^{2} \sigma}{r}+\frac{4 \pi R^{2} \sigma}{R}\right)\)

V = \(\frac{(r+R) \sigma}{\varepsilon_{0}}\)

Substituting for o, we have

Numerical Problems :

Formulae for solving numerical problems

• V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\) electric potential at a point.
• Q = CV for a capacitor.
• C = 4πε₀R for a spherical conductor.
• C =\(\frac{Q}{V}=\frac{\varepsilon_{0} A}{d}\) capacitance of a parallel plate capacitor with air as dielectric.
• C = \(\frac{K \varepsilon_{0} A}{d}\) capacitance of a parallel plate capacitor with a dielectric.
• For capacitors in senes \(\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}\) and for capacitors in parallel C_p = C₁ + C₂
  
• U = \(\frac{1}{2} \frac{Q^{2}}{c}\) = \(\frac{1}{2}\)CV² = \(\frac{1}{2}\)QV Energy stored in a capacitor.
• Energy density u = \(\frac{1}{2}\) ε₀E₂
  
• Capacitance of a parallel plate capacitor with a conducting slab of thickness t between plates is C = \(\frac{\varepsilon_{0} A}{d-t}\)
• Capacitance of a capacitor with dielectric slab of thickness t << d , C = \(\frac{\varepsilon_{0} A}{d+t\left(\frac{1}{K}-1\right)}\)
• Common potential V = \(\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}\)
• Loss of energy when two conductors are combined, U₁ – U₂ = \(\frac{C_{1} C_{2}}{2\left(C_{1}+C_{2}\right)}\left(V_{1}-V_{2}\right)^{2}\)
• If n small drops each having a charge Q, capacitance C, and potential V coalesce to form a big drop, then

1. The charge on the big drop = nQ
2. The capacitance of the big drop = n^1/3 C
3. Potential of the big drop = n^2/3 V
4. The potential energy of the big drop = n^5/3 U

Question 1.
Electric field intensity at point B due to a point charge Q kept at point A is 24 N C^-1 and the electric potential at point B due – to the same charge is 12 J C^-1. Calculate the distance AB and also the magnitude of the charge Q.
Answer:
Given E = 24 N C^-1 , V = 12 J C^-1 , r = ? and Q = ?

Using the relation
E = \(\frac { V }{ r }\)
Or
\(\frac { V }{ E }\) = \(\frac { 12 }{ 24 }\) = 0.5 m

Also V = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r}\)

Therefore 12 = 9 × 10⁹ × \(\frac { Q }{ 0.5 }\) , solving for Q

we have Q = 6.67 × 10^-10C

Question 2.
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 μC. When the potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 μC. Calculate:
(i) The potential V and the unknown capacitance C.
(ii) What will be the charge stored in the capacitor, if the voltage applied had increased by 120 V? (CBSE Delhi 2013)
Answer:
Given Q₁ = 360 μC, Q₂ = 120 μC,
V₂ = (V- 120) volt, V₁ = V

(i) We know that C = \(\frac { Q }{ V }\)

Since capacitance is same, we have
\(\frac{Q_{1}}{V_{1}}=\frac{Q_{2}}{V_{2}}\)
Or
\(\frac{360 \times 10^{-6}}{V}=\frac{120 \times 10^{-6}}{V-120}\)

Solving for V we have V= 180 volt

Also C = Q₁/ V₁ = 360 × 10^-6 / 180 = 2 × 10^-6 C = 2 pF

(ii) V= 180 + 120 = 300 V
Therefore Q = CV = 2 x 10^-5 × 300 = 600 x 10^-6 = 600 pC

Question 3.
Two identical capacitors of 12 pF each are connected in series across a 50 V battery. Calculate the electrostatic energy stored in the combination. If these were connected in parallel across the same battery, find out the value of the energy stored in this combination. (CBSE Al 2019)
Answer:
Given C = 12 pF = 12 × 10^-12 F, V= 50 V,
In series
C_s = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{12 \times 12}{12+12}\) = 6 pF

Hence energy stored
U_s = \(\frac{1}{2}\) C_sV₂ = \(\frac{1}{2}\) × 6 × 10^-12× (50)²
U_s = 7.5 × 10^-9 J

In parallel
C_P = C₁ + C₂ = 12 + 12 = 24 pF
U_P = \(\frac{1}{2}\) C_PV² = \(\frac{1}{2}\) × 24 × 10^-12 × (50)²
U_P = 3 × 10^-8 J

Question 4.
The figure shows a network of three capacitors C₁ = 2 μF; C₂ = 6 μF and C₃ = 3 μF connected across a battery of 10 V. If a charge of 6 μC is acquired by the capacitor C₃, calculate the charge acquired by C₁ (CBSE Al 2019)

Answer:
Capacitors C₂ and C₃ are connected in parallel, therefore, the net capacitance of the combination.
C₂₃ = (6 + 3) = 9 μF

Let V1, be the potential across C1 and V2 be the potential across C₂₃
Now C₁ = Q/V, or V₁ = Q/C₁ = Q/2

Also V₂ = Q/9
But V = V₁ + V₂

Solving for Q we get
Q= 16.4 μC

Question 5.
(a) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 μF capacitance.

Answer:
This can be redrawn as

Like C₂, C₃ and C₄ are in parallel,
C₂₃₄ = C₂ + C₃ + C₄ = 6 μF

Further, C₁, C₂₃₄ and C₅ are in series
\(\frac{1}{C_{\text {net }}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}=\frac{3+1+3}{6}=\frac{7}{6}\)
\(C_{\text {net }}=\frac{6}{7}\)μF

(b) If a DC source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network? (CBSE Delhi 2017)
Answer:
∴ C_net = \(\frac { 6 }{ 7 }\) μF ,

q = C_net V = \(\frac { 6 }{ 7 }\) × 10^-6 × 7 = 10^-6C

∴ Energy stored = \(\frac{1}{2}\)qV

= \(\frac{1}{2}\) × 6 × 10^-6 × 7 = 21 × 10^-6J

Question 6.
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also, find the charge drawn from the battery in each case. (CBSE Delhi 2017)
Answer:
(i) Net capacitance C_net = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}\)
= \(\frac{12 \times 12}{12+12}\)pF = 6 pF

∴ Energy stored = \(\frac{1}{2}\)C_net V²
= \(\frac{1}{2}\) × 6 × 10^-12 × (50)² = 75 × 10^-10J
q = Chargedrawn = C_net V=6 × 10^-12 × 50 = 3 × 10^-10 C

(ii) C_net=12 + 12 = 24pF
∴ Energy stored = \(\frac{1}{2}\) C_net V²
= \(\frac{1}{2}\) × 24 × 10^-12 × (50)²
= 3 × 10^-8J

Charge drawn, q = C_netV
= 24 × 10^-12 × 50
= 1200 × 10^-12
= 12 × 10^-10 C

Question 7.
In the following arrangement of capacitors, the energy stored in the 6 μF capacitor is E. Find the value of the following. (Foreign 2016)
(a) Energy stored in 12 pF capacitor.
(b) Energy stored in 3 pF capacitor.
(c) Total energy drawn from the battery.

(a) Energy stored in 6 μ capacitor is E. Capacitors 6 μF and 12 μF are connected in parallel.
So, the voltage across 6 μF capacitor
= voltage across 12 μF capacitor
= voltage across 12 μF capacitor
= V (say)

As E = \(\frac{1}{2}\) × 6 × V²
∴ V = \(\sqrt{\frac{E}{3}}\)

Similarly energy U’ stored in 12 pF capacitor
= \(\frac{1}{2}\) × 12 × V²
= \(\frac{1}{2}\) ×12 × \(\frac{E}{3}\) = 2E

U’ = 2 E

(b) Equivalent capacitance of 6 µF and 12 µF is 6 + 12 = 18 µF

Charge on 18 µF and 3 µF is same as they are in series as they are in series
∴ Q = CV= 18 × V

∴ Energy in 3 µF capacitor
U” = \(\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} \frac{(18 V)^{2}}{3}\)

U” = \(\frac{1}{2} \times \frac{18 \times 18}{3} \frac{E}{3}\)
U” = 18E

(c) Total energy drawn from battery U = E + 2E + 18E = 21E

Question 8.
Two parallel plate capacitors X and Y have the same area of plates and the same separation between them, X has air between the plates, while Y contains a dielectric medium of ε_r = 4.
(a) Calculate the capacitance of each capacitor if the equivalent capacitance of the combination is 4 μF.

C_X = \(\frac{\varepsilon_{0} A}{d}\) = C(say)
C_Y= 4\(\frac{\varepsilon_{0} A}{d}\) = 4C

EquivaLent capacitance = 4 μF
C_eq = \(\frac{C_{X} C_{Y}}{C_{X}+C_{Y}}\) = 4
= \(\frac{C \times 4 C}{C+4 C}\)
Or
= \(\frac{4C}{5}\) = 4

C = 5 μF
∴ C_X = C = 5 μF and
C_Y = 4C = 20 μF

(b) Total charge, q = C_eq V = 4 × 12 = 48μC
C_X and C_Y are in series. Hence, charge on both is 48 μC each.

∴ The potential difference across C_X,
V_X = \(\frac{q}{C_{X}}=\frac{48}{5}\) = 9.6 V0lt

The potential difference across C_Y,
V_Y = \(\frac{q}{C_{Y}}=\frac{48}{20}\) = 2.4 Volt

(c) U_X=\(\frac{1}{2}\)CXVX² ; UY = \(\frac{1}{2}\)CYVy²

∴ \(\frac{U_{x}}{U_{Y}}=\frac{5 \times(9.6)^{2}}{20 \times(2.4)^{2}}\)
= \(\frac{1}{4} \times \frac{9.6 \times 9.6}{2.4 \times 2.4}\) = 4

\(\frac{U_{x}}{U_{r}}\) = 4

Question 9.
Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C₁ and C₂ with their capacitances in the ratio 1:2 so that the energy stored in these two cases becomes the same. (CBSE Al 2016)
Answer:
(i) Let C₁ = C, ∴ C₂ = 2C
For series combination equivalent capacitance is
\(C_{s}=\frac{C_{1} \times C_{2}}{C_{1}+C_{2}}=\frac{C \times 2 C}{3 C}=\frac{2}{3} C\) …. (1)
and energy stored,
\(U_{\mathrm{s}}=\frac{1}{2} C_{s} V_{\mathrm{s}}^{2}=\frac{1}{2} \times \frac{2}{3} C V_{\mathrm{s}}^{2}\) …. (2)

For parallel combination equivalent capacitance
C_p = C + 2C = 3C ….(3)
and energy stored
∴ \(U_{D}=\frac{1}{2} 3 C \times V_{p}^{2}\) …(4)

But U_s = U_p [Given]

Question 10.
Calculate the potential difference and the energy stored in the capacitor C₂ in the circuit shown in the figure. Given potential at A is 90 V, C₁ = 20 μF, C₂ = 30 μF, and C₃= 15 μF. (CBSEAI 2015)

Answer:

C = \(\frac{60}{9}\) μF = \(\frac{20}{3}\) μF

The potential at A = 90V

∴ Charge on each capacitor
Q = C x V= \(\frac{20}{3}\) x 90 =600 μC.

∴ Charge on C₂ = 600 μC

∴ V₂ = \(\frac{Q}{C_{2}}=\frac{600}{30}\) ; V₂ = 20 V

Energy stored in C₂ = \(\frac{1}{2}\)C₂V₂²

U₂ = \(\frac{1}{2}\) × 3o × 10^-6 × 20 × 20
U₂ = 6000 × 10^-6J = 6 × 10^-3J

Class 12 Chemistry Chapter Wise Important Questions with Answers Pdf Download 2020-21: Here we are providing CBSE Important Extra Questions for Class 12 Chemistry Chapter Wise Pdf download in Hindi and English Medium. Students can get Class 12 Chemistry NCERT Solutions, Chemistry Class 12 Important Extra Questions and Answers designed by subject expert teachers.

CBSE Class 12th Chemistry Chapter Wise Important Questions with Answers Pdf

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16. Chemistry in Everyday Life Class 12 Important Questions

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Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 1 Electric Charges and Fields. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 1 Important Extra Questions Electric Charges and Fields

Electric Charges and Fields Important Extra Questions Very Short Answer Type

Question 1.
What is the value of the angle between the vectors \(\vec{p}\) and \(\vec{E}\) for which the potential energy of an electric dipole of dipole moment \(\vec{p}\), kept in an external electric field \(\vec{E}\), has maximum value.
Answer:
P.E. = –pEcos θ
P.E. is maximum when cos θ = – 1, i.e.
θ = 180°

Question 2.
Define electric field intensity at a point.
Answer:
Electric field intensity at a point is defined as the force experienced by a unit test charge placed at that point. Mathematically
we have

Question 3.
Two equal point charges separated by 1 m distance experience force of 8 N. What will be the force experienced by them, if they are held in water, at the same distance? (Given: K_water = 80) (CBSE Al 2011C)
Answer:
The force in water is given by
F_w = \(\frac{F_{\text {air }}}{K}=\frac{8}{80}\) = 0.1 N

Question 4.
A charge ‘q’ is placed at the center of a cube of side l. What is the electric flux passing through each face of the cube? (CBSE AI 2012) (CBSE Sample Paper 2019)
Answer:
Φ = q/6ε₀

Question 5.
Why do the electric field lines not form closed loops? (CBSE Al 2012C)
Answer:
It is due to the conservative nature of the electric field.

Question 6.
Two equal balls having equal positive charge ‘q’ coulomb are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two? (CBSE AI 2014)
Answer:
It decreases because force ∝= \(\frac{1}{k}\) and k > 1.

Question 7.
What is the electric flux through a cube of side l cm which encloses an electric dipole? (CBSE Delhi 2015)
Answer:
Zero

Question 8.
Why are electric field lines perpendicular at a point on an equipotential surface of a conductor? (CBSE Al 2015C)
Answer:
So that no net force acts on the charge at the equipotential surface and it remains stationary.

Question 9.
What is the amount of work done in moving a point charge Q. around a circular arc of radius ‘r’ at the centre of which another point charge ‘q’ is located? (CBSE Al 2016)
Answer:
Zero.

Question 10.
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased? (CBSE Delhi 2016)
Answer:
No change, as flux does not depend upon the size of the Gaussian surface.

Question 11.
Draw the pattern of electric field lines, when a point charge –Q is kept near an uncharged conducting plate. (CBSE Delhi 2019)
Answer:
The pattern is as shown.

Question 12.
Draw a plot showing the variation of the electric field with distance from the center of a solid conducting sphere of radius R, having a charge + Q on its surface. (CBSE Delhi 2017C)
Answer:
The plot is as shown.

Question 13.
Draw a graph to show the variation of E with perpendicular distance r from the line of charge (CBSE Delhi 2018)
A
Answer:
E = \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\)
E ∝ \(\frac{1}{r}\)

The graph between E and r is as shown.

Question 14.
Define the term ‘electric flux’. Write its S.I. unit. (CBSE Delhi 2018)
Answer:
Electric flux: It is the measure of the number of electric field lines crossing a given area normally.

Mathematically the electric flux passing through an area \(\vec{dS}\) is given by
dΦ = \(\vec{E}\) . \(\vec{dS}\)

SI unit of electric flux is Nm²C^-1 or Vm.

Question 15.
Why can the interior of a conductor have no excess charge in the static situation? (CBSE Ai 2019)
Answer:
Since the electric field inside the conductor is zero, electric flux through the closed surface is also zero. Hence by Gauss’s law, there is no net charge enclosed by the closed surface.

Question 16.
Two field lines never cross each other. Why?
Answer:
It is because at the point of intersection two perpendiculars can be drawn. Thus there will be two directions of the electric field at that point which is not possible.

Question 17.
In an electric field, an electron is kept freely. If the electron is replaced by a proton, what will be the force experienced by the proton?
Answer:
The magnitude of force will be the same but the direction will be reversed.

Question 18.
Consider the situation shown in the figure given below. What are the signs of q₁ and q₂?
Answer:

q₁ is negative and q₂ is positive.

Question 19.
In the figure given below, at which point electric field is maximum?
Answer:

The electric field is maximum at point C.

Question 20.
An uncharged insulated conductor A is brought near a charged insulated conductor B. What happens to the charge and potential of B?
Answer:
On bringing uncharged conductor A near a charged conductor B, charges are induced on A as shown in the figure below. As a result, the potential of conductor B is slightly lowered but the charge on it remains unchanged.

Question 21.
In a medium the force of attraction between two point electric charges, distance ‘d’ apart, is F. What distance apart should these be kept in the same medium so that the force between them becomes 3F?
Answer:
Let the new distance be ‘d’, since F ∝\(\frac{1}{r^{2}}\) ,
therefore \(\frac{F}{3 F}=\frac{r^{2}}{d^{2}}=\frac{1}{3} \Rightarrow r=\frac{d}{\sqrt{3}}\)

Question 22.
Find the value of an electric field that would completely balance the weight of an electron.
Answer:
mg = eE ⇒ E = \(\frac{m g}{e}\)
= \(\frac{9.1 \times 10^{-31} \times 9.8}{1.6 \times 10^{-19}}=5.57 \times 10^{-11} \mathrm{Vm}^{-1}\)

Question 23.
Two charges, one +5 µC, and the other -5 µC are placed 1 mm apart. Calculate the electric dipole moment of the system.
Answer:
p = q × 2a = 5 × 10 ^-6 × 10^-3 = 5 × 10^-9 Cm

Question 24.
Two-point charges of+3 µC each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero?
Answer:
At the mid-point of the line joining the two point charges.

Question 25.
Four charges of came magnitude and same sign are placed at the corners of a square, of each side 0.1 m. What is electric field intensity at the center of the square?
Answer:
Zero.

Question 26.
Why should the electrostatic field be zero ‘ inside a conductor? (CBSE Delhi 2012)
Answer:
Because it does not contain any charge.

Question 27.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the shell. What will be the surface charge density on the (i) inner surface, and (ii) the outer surface of the shell? (CBSE Delhi 2018)
Answer:
On inner surface

On the outer surface,

Question 28.
An arbitrary surface encloses a dipole. What is the electric flux through this surface? (NCERT Exemplar)
Answer:
Zero.

Electric Charges and Fields Important Extra Questions  Short Answer Type

Question 1.
(a) Electric field inside a conductor is zero. Explain.
(b) The electric field due to a point charge at any point near it is given as
E  =

what is the physical significance of this limit?
Answer:
(a) By Gauss theorem \(\phi \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_{0}}\). Since there is no charge inside a conductor therefore in accordance with the above equation the electric field inside the conductor is zero.
(b) It indicates that the test charge should be infinitesimally small so that it may not disturb the electric field of the source charge.

Question 2.
Define the electric line of force and give its two important properties.
Answer:
It is a line straight or curved, a tangent to which at any point gives the direction of the electric field at that point.
(a) No two field lines can cross, because at the point of intersection two tangents can be drawn giving two directions of the electric field which is not possible.
(b) The field lines are always perpendicular to the surface of a charged conductor.

Question 3.
Draw electric field lines due to (i) two similar charges, (ii) two opposite charges, separated by a small distance.
Answer:
(a) The diagram is as shown.

(b) The diagram is as shown.

Question 4.
An electric dipole is free to move in a uniform electric field. Explain what is the force and torque acting on it when it is placed
(i) parallel to the field
Answer:
When an electric dipole is placed parallel to a uniform electric field, net force, as well as net torque acting on the dipole, is zero and, thus, the dipole remains in equilibrium.

(ii) perpendicular to the field
Answer:
When the dipole is placed perpendicular to the field, two forces acting on the dipole form a couple, and hence a torque acts on it which aligns its dipole along the direction of the electric field.

Question 5.
A small metal sphere carrying charge +Q. is located at the center of a spherical cavity in a large uncharged metallic spherical shell. Write the charges on the inner and outer surfaces of the shell. Write the expression for the electric field at the point P₁(CBSE Delhi 2014C)
Answer:

Charge on inner surface – Q.
Charge on outer surface + Q,
Electric field at point P = E = k\(\frac{Q}{r_{1}^{2}}\)

Question 6.
Two-point charges q and –2q are kept ‘d’ distance apart. Find the location of the point relative to charge ‘q’ at which potential due to this system of charges is zero. (CBSE Al 2014C)
Answer:
Let the potential be zero at point P at a distance x from charge q as shown

Now potential at point P is
V = \(\frac{k q}{x}+\frac{k(-2 q)}{d+x}\) = 0

Solving for x we have
x = d

Question 7.
Two small identical electrical dipoles AB and CD, each of dipole moment ‘p’ are kept at an angle of 120° as shown in the figure. What is the resultant dipole moment of this combination? If this system is subjected to the electric field (\(\vec{E}\)) directed along +X direction, what will be the magnitude and direction of the torque acting on this? (CBSE Delhi 2011)

Answer:
The resultant dipole moment of the combi-nation is
P_R = \(\sqrt{p^{2}+p^{2}+2 p^{2} \cos 120^{\circ}}\) = p

since cos 120° = -1/2
This will make an angle of 30° with the X-axis, therefore torque acting on it is
τ=PE sin 30° = pE/2 (Along Z-direction)

Question 8.
A metallic spherical shell has an inner radius R₁ and outer radius R₂. A charge Q is placed at the center of the spherical cavity. What will be surface charge density on (i) the inner surface, and (ii) the outer surface? (NCERT Exemplar)
Answer:
The induction of charges is as shown.

Therefore surface charge density on the inner and the outer shell is on the outer surface is

Question 9.
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that the net charge inside is zero. (NCERT Exemplar)
Answer:
No, the field may be normal to the surface. However, the converse is true i.e. when the electric field everywhere on the surface be zero then the net charge inside it must be zero.

Electric Charges and Fields Important Extra Questions Long Answer Type

Question 1.
(a) State Gauss theorem in electrostatics. Using it, prove that the electric field at a point due to a uniformly charged infinite plane sheet is independent of the distance from it.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)
Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law
Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have
2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)
E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged? (C8SE Delhi 2012)
Answer:
(a) directed outwards
(b) directed inwards.

Question 2.
Use Gauss’s law to derive the expression for the electric field (\(\vec{E}\) ) due to a straight uniformly charged infinite line of charge λ Cm^-1. (CBSE Delhi 2018)
Answer:

Consider an infinitely Long, thin wire charged positively and having uniform Linear charge density λ. The symmetry of the charge distribution shows that must be perpendicular to the tine charge and directed outwards. As a result of this symmetry, we consider a Gaussian surface in the form of a cylinder with arbitrary radius r and arbitrary Length L. with its ends perpendicular to the wire as shown in the figure. Applying Gauss’s theorem to curved surface ΔA₁ and circular surface ΔA₂.

Φ EΔA1 cos 0°+ EΔA2 cos 90° = \(\frac{q}{\varepsilon_{0}}\) = \(\frac{\lambda l}{\varepsilon_{0}}\) [∵ λ = \(\frac{q}{e}\)]
Or
E . 2πrl = \(\frac{\lambda l}{\varepsilon_{0}}\) ⇒ E = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{\lambda}{r}\)

This is the expression for the electric field due to an infinitely long thin wire.
The graph is as shown.

Question 3.
Obtain the expression for the potential energy of an electric dipole placed with its axis at an angle (θ) to an external electric field (E). What is the minimum value of the potential energy? (CBSE 2019C)
Answer:
The torque x acting on an electric dipole of dipole moment p placed in a uniform electric field E is given by:
τ = pEsinθ …(i)
where θ is the angle made by the dipole with the electric field E. The torque tends to align the dipole along the direction of the field. If the dipole is rotated through a small angle dθ against the torque, work has to be done, which is stored in the form of the potential energy of the dipole.

Work done in rotating the dipole through the angle dθ against the torque τ is given by dw = τ dθ = pE sin θ dθ

If the dipole is rotated from θ₁ to θ₂, then
Total work is done,

W = – pE(cos θ₂ – cos θ₁)

This work done is stored as potential energy U.

U = – pE(cos θ₂ – cos θ₁)
If the dipole is rotated from θ₁ = \(\frac{\pi}{2}\) to θ₂ =θ, then

U = – pE(cos θ – cos \(\frac{\pi}{2}\) )
= – pE(cos θ – 0)

U = – pE cos θ
minimum value of potential energy
U = – pE When θ = 0°

Question 4.
Why does the electric field inside a dielectric decrease when it is placed in an external electric field?
Answer:
When a dielectric is placed in an electric field (E₀), it gets polarised, i.e. within the dielectric, an electric field (E) is induced in a direction opposite to that of the external field. Therefore, the net field within the dielectric decreases to \(\vec{E}_{0}\) – \(\vec{E}\)

Question 5.
Two-point charges +q and -2q are placed at the vertices ‘B’ and ‘C’ of an equilateral triangle ABC of side ‘a’ as given in the figure. Obtain the expression for (i) the magnitude and (ii) the direction of the resultant electric field at vertex A due to these two charges. (CBSE Al 2014 C)

Answer:
The electric fields due to the two charges placed at B and C are inclined at an angle θ = 120° as shown
Now in magnitude, we have

E_B = k\(\frac{q}{a^{2}}\) and
E_C = k\(\frac{2q}{a^{2}}\) = 2 E_B

Hence E = \(\sqrt{E_{B}^{2}+E_{c}^{2}+2 E_{B} E_{c} \cos \theta}\)
Or E = \(\sqrt{E_{B}^{2}+\left(2 E_{B}\right)^{2}+2 E_{B}\left(2 E_{B}\right) \cos 120^{\circ}}\)

On Solving we have
E = \(\sqrt{3} E_{B}\) = \(\sqrt{3} \frac{k q}{a^{2}}\)

Direction

Therefore B = 90°, the resultant is inclined at an angle of 90° with EB.

Question 6.
Four-point charges Q, q, Q, and q are placed at the corners of a square of side ‘a’ as shown in the figure.
Find the
(a) the resultant electric force on a charge Q and
(b) the potential energy of this system. (CBSEAI, Delhi 2018)
Answer:
(a) Let us find the force on charge Q at point C.

Force due to charge Q placed at point A is
F_A = k\(\frac{Q^{2}}{(a \sqrt{2})^{2}}\) =k\(\frac{Q^{2}}{2 a^{2}}\) along AC

Force due to the charge q placed at D
F_D = k\(\frac{q Q}{a^{2}}\) alongDA

Force due to the charge q placed at B
F_B = k \(\frac{q Q}{a^{2}}\) along BC

The resuLtant of F_D and F_B is
F_BD = K\(\frac{q Q \sqrt{2}}{a^{2}}\) along AC

∴ net force of charge Q placed at point C is

(b) PotentiaL energy of the system

Question 7.
A charge +Q is uniformly distributed within a sphere of radius R. Find the electric field, distant r from the center of the sphere where: (1) r < R and (2) r > R. (CBSEAI 2016C)
Answer:
For a solid sphere p = \( \frac{q}{\frac{4}{3} \pi R^{3}}\) = \(\frac{q}{\text { volume }}\)

Case 1. 0 < r < R The point Lies within the sphere.

Using Gauss’s theorem
Let Q’ be the charge encLosed by Gaussian’s surface of radius r < R.

E(4πr²)=\(\frac{Q^{\prime}}{\varepsilon_{0}}=\frac{Q^{\prime}}{4 \pi \varepsilon_{0} r^{2}}\)

From (i) and (ii)
E = \(\frac{Q r^{3}}{4 \pi \varepsilon_{0} r^{2} R^{3}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q r}{R^{3}}\)

Case 2. For r > R
\(\oint \vec{E} \cdot \overrightarrow{d s}\) = \(\frac{q_{\text {enclosed }}}{\varepsilon_{0}}\)

E(4πr²) = \(\frac{Q}{\varepsilon_{0}}\)
E = \(\frac{Q}{4 \pi \varepsilon_{0} r^{2}}\)

Question 8.
(a) An electric dipole is kept first to the left and then to the right of a negatively charged infinite plane sheet having a uniform surface charge density. The arrows p₁ and p₂ show the directions of its electric dipole moment in the two cases. Identify for each case, whether the dipole is in stable or unstable equilibrium. Justify each answer.
Answer:
p₁: stable equilibrium, p₂: unstable equilibrium. The electric field, on either side, is directed towards the negatively charged sheet and its magnitude is independent of the distance of the field point from the sheet. For position P₁ dipole moment and electric field are parallel. For position p₂, they are antiparallel.

(b) Next, the dipole is kept in a similar way (as shown), near an infinitely long straight wire having uniform negative linear charge density. Will the dipole be in equilibrium at these two positions? Justify your answer. (CBSE Sample Paper 2018-2019)

Answer:
The dipole will not be in equilibrium in any of the two positions.
The electric field due to an infinite straight charged wire is non-uniform (E ∝ 1/r).
Hence there will be a net non-zero force on the dipole in each case.

Question 9.
Two large parallel plane sheets have uniform charge densities +σ and -σ. Determine the electric field (i) between the sheets, and (ii) outside the sheets. (CBSE Delhi 2018)
Answer:
Let us consider two parallel planes charged conductors A and B carrying +ve and -ve charge density σ (charge per unit area). According to Gauss’ theorem, the electric intensity at P due to the charge on sheet A is

EA = \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A and B)

The electric field at P due to charge on sheet B is
E= \(\frac{\sigma}{2 \varepsilon_{0}}\) (From A to B)

The electric field at P is
E = EA + EB
= \(\frac{\sigma}{\varepsilon_{0}}\)

Question 10.
Define electric flux and write its SI unit. The electric field components in the figure shown are : E_x = αx, E_y = 0, E_z = 0 where α = 100N/cm. Calculate the charge within the cube, assuming a = 0.1 m. (CBSE Sample Paper 2019)
Or
An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10⁴ N/C (figure a)

(a) Calculate the time it takes to fall through this distance starting from rest.

(b) If the direction of the field is reversed (figure b) keeping its magnitude unchanged, calculate the time taken by a proton to fall through this distance starting from rest. (CBSE Delhi 2018C)
Answer:
Electric Flux is the dot product of the electric field and area vector.
Φ = \(\oint \vec{E} \cdot \overrightarrow{d s}\)

SI Unit: Nm²/C or Vm

For a given case
Φ = Φ₁ – Φ₂ = [Ex(atx= 2a) – Ex (atx = a)]a²
= [α(2a)-α(a)]a² = αa³
= 10⁴ × (0.1 )³ = 10 Nm²/C
But
Φ = \(\frac{q}{\varepsilon_{0}}\)

∴ q = ε₀Φ = 8.854 × 10^-12 × 10 C = 8.54 pC
Or
We have
F = qE

Acceleration, a = \(\frac{q E}{m}\)

Also
s = \(\frac { 1 }{ 2 }\) at² [u = 0]
∴ t = \(\sqrt{\frac{2 s}{a}}\)

(i) For the electron
a = \(\frac { eE}{ m }\)

∴ t = \(\sqrt{\frac{2 s m}{e E}}\)

∴ t = \(\sqrt{\frac{3 \times 10^{-2} \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^{4}}}\) = 2.92 ns

(ii) for proton
t = \(\sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2 \times 10^{4}}}\)
= -.125 μs

Question 11.
What will be the total flux through the faces of the cube (figure) with the side of length ‘a’ if a charge q is placed at
(a) A: a corner of the cube.
(b) B: mid-point of an edge of the cube.
(c) C: center of the face of the cube.
(d) D: mid-point of B and C. {NCERJ Exemplar)

Answer:
(a) The charge wilt is shared by eight cubes if it has to be enclosed. Therefore the flux through the cube will be one-eighth of the total flux. Φ = q/8ε₀
(b) The charge will be shared by four cubes if it has to be enclosed. Therefore the flux through the cube will be one-fourth of the total flux. Φ = q/ 4ε₀
(c) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀
(d) The charge will be shared by two cubes if it has to be enclosed. Therefore the flux through the cube will be one-half of the total flux. Φ = q/ 2ε₀

Question 12.
Two charges q and -3q are placed fixed on the x-axis separated by distance ‘d’. Where should a third charge 2q be placed such that it will not experience any force? (NCERT Exemplar)
Answer:
The situation is as shown.

Let the charge 2q be placed at a distance ‘x’ from charge q. For the charge 2q to experience zero force we have
\(\frac{2 q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{6 q^{2}}{4 \pi \varepsilon_{0}(d+x)^{2}}\)
(d + x)² = 3x²

Solving for x we have
x = \(\frac{d}{2} \pm \frac{\sqrt{3} d}{2}\)
(-ve sign would be between q and -3q and hence is unacceptable.)

Therefore, we have
x = \(\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3})\) to the left of q.

Question 13.
(a) State Gauss’s law. Use it to deduce the expression for the electric field due to a uniformly charged thin spherical shell at points (i) inside and (ii) outside the shell. Plot a graph showing the variation of the electric field as a function of r > R and r < R. (r being the distance from the center of the shell) (CBSE Delhi 2011, Al 2013)
Answer:
Gauss’s law states that the net outward flux through any closed surface is equal to 1 /ε₀ times the charge enclosed by the closed surface.

Consider a thin spherical shell of radius R and center at O. Let σ be the uniform surface charge density (charge per unit surface area) and q be the total charge on it. The charge distribution is spherically symmetric. Three cases arise

Case 1: at a point outside the spherical shell
In order to find the electric field at a point P outside the shell let us consider a Gaussian surface in the form of a sphere of radius r (r >>R).

By symmetry, we find that the electric field acts radially outwards and has a normal component at alt points on the Gaussian sphere. Therefore by definition of electric flux we have

Φ = E × A, where A is the surface area of the Gaussian sphere therefore
Φ = E × 4πr² …(1)

But by Gauss’s law
Φ = \(\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}=\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\) … (2)

from equations (1) and (2) it follows that
E × 4πr² = \(\frac{Q}{\varepsilon_{0}}\) Or E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}\)

E × 4πr² = \(\frac{\sigma \times 4 \pi R^{2}}{\varepsilon_{0}}\)

Ē = \(\frac{\sigma R^{2}}{\varepsilon_{0} r^{2}}\) … (3)

It follows that the electric field due to a spherical shell outside it is same as that due to a point charge. Therefore for points Lying outside the spherical shell the shell behaves as if the entire charge is concentrated at the centre.

Case 2: at a point inside the spherical shell
In this case, the Gaussian surface Lies inside the shell. Since no charge is enclosed In this surface therefore we have

E × 4πr²=q₀/ε₀ [∵ q=0]
Therefore E = 0

(b) Two identical metallic spheres A and B having charges +4Q. and -10 Q are kept a certain distance apart. A third identical uncharged sphere C is first placed in contact with sphere A and then with sphere B. Spheres A and B are then brought in contact and then separated. Find the charges on the spheres A and B. (CBSEAI 2011C)
Answer:
The initial charge on the sphere A = + 4 Q.
The initial charge on the sphere B = -10 Q.

Since all the three spheres are identical, i.e. they have the same capacity, therefore when uncharged sphere C is placed in contact with A, the total charge is equally shared between them.

Charge on C after contact with A
= \(\frac{0+4 Q}{2}\) = 2Q

Charge on A after contact with C is 2Q.
When sphere C carrying a charge 2Q is placed in contact with B, again charges are equally shared between C and B equally.

Charge on C after it is in contact with B
= \(\frac{2 Q-10 Q}{2}\) = -4Q

Now when sphere A with a charge of 2Q. is placed in contact with B, with charge -4Q.

Charge are again shared
∴ charge on A or B = \(\frac{2 Q-4 Q}{2}\) = -Q.

Question 14.
(a) Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
Answer:
It is defined as the product of the magnitude of either of the two charges and the distance between them.
For derivation see sol. 9(a) of LA-II.

(b) Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero. (CBSE Al 2013)
Answer:
The diagram is as shown

The zero potential points lie on the equatorial line.

Question 15.
(a) Using Gauss’s law obtain expressions for the electric field (i) inside, and (ii) outside a positively charged spherical shell.
(c) A Square plane sheet of side 10 cm is inclined at an angle of 30° with the direction of a uniform electric field of 200 NC“1. Calculate the electric flux passing through the sheet. (CBSE 2019C)
Answer:
(a) Spherical Shell
Consider a spherical shell of radius R. Let q be a charge on the shell. Let us find the electric field at a point P at a distance r from the center 0 of the spherical shell.

Case (i): When point P lies inside the spherical shell
From the point, P draws a Gaussian surface which will be a sphere of radius r.

From the Gauss’s Theorem,
\(\oint_{s} \vec{E} \cdot \vec{d} S=\frac{0}{\varepsilon_{0}}\) [∵ No charge exists inside the spherical shell]

Or
E = 0
i. e. electric field inside the charged spherical shell is zero.

Case (ii): When point P is lying outside the shell (i.e. r > R)
From point P, draw a Gaussian surface which will be a spherical shell of radius r. Let dS be a small area element on the Gaussian surface P.


Or

i. e. the electric field outside the spherical shell behaves as if the whole charge is concentrated at the center of the spherical shell.

(b) Show graphically variation of the electric field as a function of the distance r from the center of the sphere.
Answer:
Variation of electric field E with distance
The given figure shows the variation of an electric field with distance from the center of the charged spherical shell.

(c) Here E = 200 N/C, S = 0.1 × 0.1 = 0.01 m²
And θ = 90° – 30° = 60°

The electric flux linked with the square sheet
Φ = E S cos 60°
= 200 × 0.01 × \(\frac { 1 }{ 2 }\) = 1.0 Nm² C^-1

Question 16.
A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at a large distance from the ring, it behaves like a point charge. (CBSE Delhi 2016)
Answer:
Consider a uniformly charged ring of radius ‘a’. Let the total charge on the ring be Q, Let us find the electric field on the axis of the ring at point P distance x from the center of the ring. Consider a segment of charge dQ as shown in the figure.

The magnitude of etectñc field at P due to the segment is
dE=k\(\frac{d Q}{r^{2}}\) …(1)

This field can be resolved into its components: x component dE_x = dE cos α an along the axis of the ring and y component dE perpendicular to the axis. Since these perpendicular components, due to alt the charge segments, are equal and opposite, therefore they cancel out each other. From the diagram we have r = \(\sqrt{x^{2}+a^{2}}\) and cos α = x/r, therefore we have
…(2)

In this case, all the segments of the ring give the same contribution to the field at P since they are all equidistant from this point. Thus we can easily sum over all segments to get the total electric field at point P

If the point of observation is far away, i.e. x >> a, then E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{x^{2}}\). This is the same as that for a point charge. Thus at far-off axial points, a charged ring behaves as if a point charge is situated at the center of the ring.

Question 17.
Two thin concentric and coplanar spherical shells, of radii ‘a’ and ‘b’ (b > a), carry charges, q, and Q respectively. Find the magnitude of the electric field, at a point distant x, from their common center for
(i) O < x < a
(ii) a ≤ x < b
(iii) b ≤ x < ∞ (CBSE Delhi 2016C)
Answer:
The diagram is as shown.

(i) For 0 < x < a
Point Lies inside both the spherical shells.

∴ charge enclosed = 0
Hence, E(x) = 0

(ii) For a < x < b
Point is outside the spherical shell of radius ‘a’ but inside the spherical Shell of radius ‘b’.
Therefore

(iii) For b ≤ x< ∞
The point is outside of both the spherical shells. The total effect we charge at the center equals (Q + q).

Question 18.
(a) Define electric flux. Is It a scalar or a vector quantity?

A point charge q is at a distance of d/2 directly above the center of a square of side d, as shown In the figure. Us. Gauss’s law to obtain the expression for the electric flux through the square.
Answer:
The electric flux through a given surface is defined as the dot product of the electric field and area vector over that surface. It is a scalar quantity.

Constructing a cube of side ‘d’ so that charge ‘q’ gets placed within this cube (Gaussian surface)
According to Gauss’s law the electric flux
Φ = \(\frac{q}{\varepsilon_{0}}\)

This is the total flux through all the six faces of the cube.

Hence electric flux through the square (one face of the cube)
Φ = \(\frac{1}{6} \frac{q}{\varepsilon_{0}}\)

(b) If the point charge is now moved to a distance ‘d’ from the center of the square and the side of the square is doubled, explain how the electric flux will be affected. (CBSE Al, Delhi 2018)
Answer:
If the charge is moved to a distance ‘d’ and the side of the square is doubled the cube will be constructed to have a side 2d but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before.

Question 19.
(a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
Answer:
The required graph is

(b) Find the work done in bringing a charge q from perpendicular distance r₁ to r₂ (r₂ > r₁). (CBSEAI, Delhi 2018)
Answer:
Work done in moving the charge “q”. through a smaLL displacement ‘dr’

dW = \(\vec{F}\) . \(\vec{dr}\) =q\(\vec{E}\) .\(\vec{dr}\)
dW = qE dr cos 0 = qEdr
dW= q x \(\frac{\lambda}{2 \pi \varepsilon_{0} r}\) dr

Hence work done in moving the charge from r₁ to r₂ (r₂ > r₁)

Question 20.
Derive an expression for the torque acting on an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\). Write the direction along which the torque acts.
OR
Derive an expression for the electric field at a point on the axis of an electric dipole of dipole moment \(\vec{p}\). Also, write its expression when the distance r >> a the length ‘a’ of the dipole. (CBSEAI 2019)
Answer:
Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and E_B be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180⁰ Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

Question 21.
(a) Derive the expression for the electric field at a point on the equatorial line of an electric dipole. (CBSE 2019C)
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let E_A and E_B be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A cos θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_Acosθ + E_Bcosθ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Discuss the orientation of the dipole in (a) stable, (b) unstable equilibrium in a uniform electric field. (CBSE Delhi 2017)
Answer:
For stable equilibrium \(\vec{P}\) is along \(\vec{E}\)
θ = 0°, τ = PE sin 0°, Torque alligns the dipole in the direction of field

For unstable equilibrium \(\vec{P}\) is antiparallel to \(\vec{E}\)
∵ θ = 180°, τ = PE sin 180° = 0, Torque alligns the dipole in a direction opposite to \(\vec{E}\).

Question 22.
(a) An electric dipole of dipole moment \(\vec{p}\) consists of point charges +q and -q separated by a distance 2a apart. Deduce the expression for the electric field due to the dipole at a distance x from the center of the dipole on its axial line in terms of the dipole moment. Hence show that in

the limit r >> a, \(\vec{E}\) → \(\frac{2 \vec{p}}{4 \pi \varepsilon_{0} r^{3}}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 1800 Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Given the electric field in the region E = 2xî, find the net electric flux through the cube and the charge enclosed by it. (CBSE Delhi 2015)

Answer:
Since the electric field has only an x component, for faces perpendicular to the x-direction, the angle between E and ΔS is ± π/2.

Therefore, the flux Φ = E.ΔS is separately zero for each face of the cube except the two faces along the X-axis.

Now the magnitude of the electric field at the left face is
E_L= 0 (x = 0 at the left face)

The magnitude of the electric field at the right face is
E_R = 2x = 2a (x = a at the right face)

The corresponding fluxes are
Φ_L = E_L.ΔS = ΔS (\(\vec{E}_{L} \cdot \hat{n}_{\mathrm{L}}\)) = E_LΔS COS θ
= – E_L ΔS = 0, since θ =180°

Φ_R = E_R.ΔS = E_RΔS cos θ = E ΔS = (2a)a², since θ = 0°

Net flux through the cube
Φ = -Φ_R + Φ_L = 2a³ -0 = 2a³

We can use Gauss’s law to find the total charge q inside the cube.
We have Φ = q/ε₀ or q = Φε₀. Therefore, q = 2a³ × 8.854 × 10^-12C

Question 23.
(a) Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
Answer:
Consider an electric dipole consisting of charges -q and +q separated by a distance 2a as shown in the figure. Let the point of observation P lie on the. right bisector of the dipole AB at a distance r from its midpoint 0. Let EA and EB be the electric field intensities at point P due to charges at A and B, respectively.

The two electric fields have magnitudes

in the direction of AP

in the direction of PB.

The two fields are equal in magnitude but have different directions. Resolving the two fields E_A and E_B into their rectangular components, i.e. perpendicular to and parallel to AB. The components perpendicular to AB, i.e E_A sinG and E_B sinG being equal and opposite cancel out each other while the components parallel to AB, i.e. E_A COS θ and E_B cos θ being in the same direction add up as shown in the figure. Hence the resultant electric field at point P is given by

E = E_A cos θ + E_B cos θ

∵ cos θ = \(\frac{a}{\left(r^{2}+a^{2}\right)^{1 / 2}}\) and q 2a = p

For a short dipole r²,>>a² therefore
E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\)

(b) Two identical point charges, q each, are kept 2 m apart in the air. A third point charge Q of unknown magnitude and sign is placed on the line joining the charges such that the system remains in equilibrium. Find the position and nature of Q. (CBSE Delhi 2019)
Answer:
The third charge Q. wilt is in equilibrium if it experiences zero net force. Let it be placed at a distance x meter from the charge q.

Solving for x, we have x= 1 m

For the equilibrium of charges “q”, the nature of charge Q must be opposite to the nature of charge q and should be placed at the center of two charges.

Question 24.
(a) Define electric flux. Write its S.I. unit.
Answer:
It is defined as the total number of electric field lines crossing a given area. The electric flux can be found by multiplying the component of the electric field in the direction of the area vector (or perpendicular to the area) with the area of the closed surface. It is measured in Nm²C-^-1.

(b) A small metal sphere carrying charge +Q is located at the center of a spherical cavity inside a large uncharged metallic spherical shell as shown in the figure. Use Gauss’s law to find the expressions for the electric field at points P₁ and P₂.

Answer:
For point P, using Gauss law we have

Since E and dA are in the same direction therefore we have E = \(\frac{Q}{\varepsilon_{0} A}\)

Point P₂ lies inside the metal, therefore the Gaussian surface drawn at P₂ does not include a charge, hence the electric field at P₂ is zero.

(c) Draw the pattern of electric field lines in this arrangement. (CBSEAI 2012C)
Answer:
The electric field lines are as shown.

Question 25.
(a) Deduce the expression for the torque acting on a dipole of dipole moment \(\vec{P}\) in the presence of a uniform electric field \(\vec{E}\)
Answer:

Consider an electric dipole consisting of charges -q and +q and dipole length d is placed in a uniform electric field E as shown in the figure.

Let the dipole moment makes an angle θ with the direction of the electric field. The two charges experience force qE each.

These forces are equal, parallel, and opposite. Therefore, the net force acting on the dipole is
F_n = qE – qE = 0

Thus the net force acting on the dipole is zero.

These two forces constitute a couple. This applies a torque on the dipole given by
τ = either force × arm of the couple
τ = qE × d sin θ, where d sin θ is the arm of the couple.
τ = q d Esin θ, where p = qd, dipole moment
τ = pE sin θ.

Consider an electric dipole consisting of -q and +q charges separated by a distance 2a as shown in the figure. Let P be the point of observation on the axial Line where the electric field has to be found. Let it be at a distance r from the center O of the dipole. Let us suppose that the dipole is placed in a vacuum.

Let E and EB be the electric fields at point P due to the charges at A and B respectively. Therefore,

The two fields at P are in opposite directions. Thus the resultant electric field at P is given by
E = \(\sqrt{E_{A}^{2}+E_{B}^{2}-2 E_{A} E_{B} \cos \theta}=\sqrt{\left(E_{B}-E_{A}\right)^{2}}\) = E_B – E_A

since θ = 180° Therefore, the resultant electric field is

Solving we have

where p = q 2a.
If the dipole is short then r >> a, therefore, ‘a’ is neglected as compared to r, hence

This field is in the direction of the dipole moment.

(b) Consider two hollow concentric spheres S₁ and S₂ and enclosing charges 2Q and 4Q. respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere S₁ change if a medium of dielectric constant ‘er’ is introduced in the space inside S1 in place of air? Deduce the necessary expression. (CBSE Al 2014)

Answer:
Φ₁ = \(\frac{2 Q}{\varepsilon_{0}}\) and Φ2 = \(\frac{2 Q+4 Q}{\varepsilon_{0}}=\frac{6 Q}{\varepsilon_{0}}\)

Hence ratio

When a dielectric of dielectric constant εr is introduced then we have
Φ₁ = \(\frac{2 Q}{\varepsilon}=\frac{2 Q}{\varepsilon_{r} \varepsilon_{0}}\)

Question 26.
(a) Use Gauss’s theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density σ.
Answer:
It states, “The net electric flux through any Gaussian surface is equal to \(\frac{1}{\varepsilon_{0}}\) times the net electric charge enclosed by the surface.

Mathematically, Φ =\( \phi \vec{E} \cdot d \vec{A}=\frac{q_{i n}}{\varepsilon_{0}}\)

Consider an infinite plane sheet of charge. Let a be the uniform surface charge density, i.e. the charge per unit surface area. From symmetry, we find that the electric field must be perpendicular to the plane of the sheet and that the direction of E on one side of the plane must be opposite to its direction on the other side as shown in the figure below. In such a case let us choose a Gaussian surface in the form of a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.

The charged sheet passes through the middle of the cylinder’s length so that the cylinder’s ends are equidistant from the sheet. The electric field has a normal component at each end of the cylinder and no normal component along the curved surface of the cylinder. As a result, the electric flux is linked with only the ends and not the curved surface.

Therefore by the definition of eLectric flux, the flux Linked with the Gaussian surface is given by
Φ = \(\oint_{A} \vec{E} \cdot \vec{\Delta} A\)

Φ = E_A + E_A = 2E_A … (1)

But by Gauss’s Law

Φ = \(\frac{q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}\) [∵ q = σA] … (2)

From equations (1) and (2), we have

2E_A = \(\frac{\sigma A}{\varepsilon_{0}}\) … (3)

E = \(\frac{\sigma}{2 \varepsilon_{0}}\) …. (4)

This gives the electric field due to an infinite plane sheet of charge which is independent of the distance from the sheet.

(b) An infinitely large thin plane sheet has a uniform surface charge density +σ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet. (CBSE Al 2017)
Answer:
Work done

Numerical Problems:

Formulae for solving numerical problems

• q = ±ne
• F = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}\)
• Fmed = \(\frac{1}{4 \pi \varepsilon_{0} K} \frac{q_{1} q_{2}}{r^{2}}\) where K is dielectric constant.
• 
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\) for an electric dipole on its axial line.
• E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{p}{r^{3}}\) for an electric dipole on its equitorial line.
• Torque on an electric dipole in a uniform electric field, τ = PE sin θ.
• U = -pE cos θ, the potential energy of an electric dipole.
• Φ = \(\oint \vec{E} \cdot d \vec{A}=\frac{q_{\text {in }}}{\varepsilon_{0}}\)
• E = \(\frac{\sigma}{2 \varepsilon_{0}}\) , electric field due to an infinite plane sheet of charge.

Question 1.
An electric dipole of length 2 cm is placed with its axis making an angle of 60° to a uniform electric field of 105 NC^-1. If it experiences a torque of 8\(\sqrt{3}\) Nm, calculate the (i) magnitude of the charge on the dipole and (ii) potential energy of the dipole.
Answer:
(i) Using
q = τ / L E sin Φ

we have
q = 8N\(\sqrt{3}\) × 2 / 2 × 10^-2 × 10⁵ × \(\sqrt{3}\) = 8 × 10^-3 C

(ii) Using U = -pE cos θ, we have
U = – 8 × 10^-3 × 0.02 × 10⁵ × 0.5 = – 8 J

Question 2.
A charge of 17.7 × 10^-4 C is distributed uniformly over a large sheet of area 200 m². Calculate the electric field intensity at a distance of 20 cm from it in the air.
Answer:
Given q = 17.7 × 10^-4C, A = 200 m², r = 20cm = 0.2 m

Using the relation

Question 3.
Two fixed point charges +4e and +e units are separated by a distance ‘a’. Where should the third point charge be placed for it to be in equilibrium?
Answer:
The third charge will be in equilibrium if it experiences zero net force. Let it be placed at a distance x from the charge 4e.

4(a – x)² = x² solving for x we have
2 (a – x) = x or x = a/3

Question 4.
An electric field along the x-axis is given by \(\vec{E}\) = 100 îN/C for x > 0 and \(\vec{E}\) = -100 îN/C for x < 0. A right circular cylinder of length 20 cm and radius 5 cm lies parallel to the x-axis with its centre at the origin and one face at x = +10 cm, the other face at x = -10 cm. Calculate the net outward flux through the cylinder. (CBSE 2019C)

r = 5 cm = 0.05 m
Φ = ?

Net outward flux through the cylinder
Φ = Φ₁ + Φ₂ + Φ₃
= \(\vec{E} \cdot \overrightarrow{d s}_{1}+\vec{E} \cdot \overrightarrow{d s}_{2}+\vec{E} \cdot \overrightarrow{d s}_{3}\)

= Eds₁ cos 180° + Eds₂ cos 90° + Eds3 cos 0°
= – Eds₁ + Eds₂ (0) + Eds cos 0°
= – (-100) ds + 100 ds
= (100 +100) ds
= 200 × πr² = 200 × 3.14 × (0.05)²
= 1.57 Nm² C^-1

Question 5.
A hollow cylindrical box of length 1m and area of cross-section 25 cm² is placed in a three-dimensional coordinate system as shown in the figure. The electric field
in the region is given by \(\vec{E}\) = 50x î, where E is in NC^-1 and x is in meters. Find

(i) Net flux through the cylinder.
Answer:
We can see from the figure that on the left face E and dS are antiparallel. Therefore, the flux is
ΦL = \(\vec{E}\).\(\vec{dS}\) = E dS cos 180°
= -50 × 1 × 25 × 10^-4 = -0.125 N m² C^-1

On the right face, E and dS are parallel and therefore
ΦR = \(\vec{E}\).\(\vec{ds}\) = EdS cos 0°
= 50 × 2 × 25 × 10^-4 = 0.250 N m² C^-1

Therefore net flux is – 0.125 + 0.250 = + 0.125 N m² C^-1

(ii) Charge enclosed by the cylinder. (CBSE Delhi 2013)

Answer:
q = Φε₀ = 0.125 × 8.854 × 10^-12 = 1.1 × 10^-12C

Question 6.
(i) Define electric flux. Write its SI units,
Answer:
(i) It is defined as the number of electric field lines crossing a unit area perpendicular to the given area. It is measured in N m² C^-1

(ii) Consider a uniform electric field \(\vec{E}\) = 5 × 10³î NC^-1. Calculate the flux of this field through a square surface of area 12 cm² when
(a) its plane is parallel to the Y – Z plane, and
(b) the normal to its plane makes a 60° angle with the x-axis. (CBSE Delhi 2013C)
Answer:
Given \(\vec{E}\) = 5 × 10³î N C^-1, A = 12 cm² = 12 × 10^-4 m²

(a) Here \(\vec{E}\) = 5 × 103î N C^-1
Area of square = 12 × 10^-4 m²

The plane of surface area being parallel to YZ plane, hence
A = 12 × 10^-4 î m²

Electric flux of the field
Φ = \(\vec{E}\) \(\vec{A}\) = (5 × 10³ î). (12 × 10^-4î) = 6 N C^-1 m²
(b) When normal to the plane of surface area makes an angle of 60° with the X-axis, the flux is given by
Φ = EA cos 0 = 5 × 10³ × 12 × 10^-4 × 0.5
= 3 NC^-1 m²

Important Questions for Class 12 Physics Chapter Wise with Solution Pdf Download 2020-21: Here we are providing CBSE Important Extra Questions for Class 12 Physics Chapter Wise Pdf download in Hindi and English Medium. Students can get Class 12 Physics NCERT Solutions, Physics Class 12 Important Extra Questions and Answers designed by subject expert teachers.

CBSE Class 12th Physics Important Extra Questions and Answers Chapter Wise Pdf

1. Class 12 Physics Chapter 1 Important Questions Electric Charges and Fields
2. Class 12 Physics Chapter 2 Important Questions Electrostatic Potential and Capacitance
3. Current Electricity Class 12 Important Questions
4. Moving Charges and Magnetism Class 12 Physics Important Questions
5. Magnetism and Matter Physics Class 12 Important Questions
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