Electric Charges and Fields Class 12 MCQs Questions with Answers
Class 12 Physics Chapter 1 MCQ Electric Charges and Fields Question 1.
Plastic rod rubbed with fur and glass rod rubbed with silk ………..
(A) repel each other
(B) mix up with each other
(C) attract each other
(D) None of the above
Answer:
(C) attract each other
Explanation:
Rubbing a rod with certain materials will cause the rod to become charged. A plastic rod when rubbed with fur becomes negatively charged and a glass rod when rubbed with silk becomes positively charged.
Chapter 1 Physics Class 12 MCQ Question 2.
Electric charge between two bodies can be produced by ……………..
(A) sticking
(B) rubbing
(C) oiling
(D) passing AC current
Answer:
(B) rubbing
Explanation:
The tribo-electric effect is a type of contact electrification on which certain materials become electrically charged after they come into frictional contact with a different material.
Physics Class 12 Chapter 1 MCQ Question 3.
Electric charges under action of electric forces is called ……………
(C) electric field
(D) electric field lines
Answer:
Option (A) is correct.
Explanation:
Coulomb force, also called electrostatic force or Coulomb interaction, attraction or repulsion of particles or objects because of their electric charge.
Electric Charges And Fields Class 12 MCQ Question 4.
Law stating that “force is directly proportional to product of charges and inversely proportional to square of separation between them” is called ……………
(A) Newton’s law.
(B) Coulomb’s law
(C) Gauss’s law
(D) Ohm’s law
Answer:
(B) Coulomb’s law
Explanation:
Coulomb’s law states that : The magnitude of the electrostatic force of attraction or repulsion between two-point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distances between them.
Class 12 Physics Chapter 1 MCQ Questions And Answers Question 5.
In given figure, two positive charges q2 and q3 fixed along the y axis, exert a net electric force in the + x direction on a charge q1 fixed along the x axis. If a positive charge Q is added at (x, 0), the force on q1
(A) shall increase along the positive z-axis.
(B) shall decrease along the positive z-axis.
(C) shall point along the negative z-axis.
(D) shall increase but the direction changes because of the intersection of Q with q2 and q3
Answer:
(A) shall increase along the positive z-axis.
Explanation:
Net force on charge q1 by other charges q2 and q3 is along the + z-direction, so nature of force between qt and q2 and q1 and q3 is attractive. This is possible when charge q2 is negative. Now, if a positive charge Q is placed at (z, 0), then, the force on qt will increase. The direction will be along positive z-axis.
Physics Chapter 1 Class 12 MCQ Question 6.
The magnitude of electric force, F is …………..
(A) directly proportional to the multiplication of both charges.
(B) directly proportional to the distance between both charges.
(C) directly proportional to the square of the distance between both charges.
(D) constant.
Answer:
(A) directly proportional to the multiplication of both charges.
Explanation:
The magnitude of the electric force F is directly proportional to the amount of an electric charge, qy multiplied by the other, qy and inversely proportional to the square of the distance V between their centres.
Ch 1 Physics Class 12 MCQ Question 7.
A body is negatively charged means ………….
(A) It has only negative charges.
(B) Positive charges have been neutralized by negative charges.
(C) The quantity of negative charge present is more than the quantity positive charge present.
(D) The positive are displaced from their original positions.
Answer:
(C) The quantity of negative charge present is more than the quantity positive charge present.
Explanation:
When a neutral body gains electrons, it becomes negatively charged. It means that the quantity of negative charges present is more than the quantity of positive charge present.
MCQ Of Chapter 1 Physics Class 12 Question 8.
When a body is charged by conduction, its mass ………..
(A) remains same.
(B) increases.
(C) decreases.
(D) increase or decrease.
Answer:
(D) increase or decrease.
Explanation:
When a body is negatively charged by conduction, it gains electrons. Hence, its mass increases. When a body is positively charged by conduction, it loses electrons. Hence, its mass decreases.
MCQ Physics Class 12 Chapter 1 Question 9.
A point positive charge is brought near an isolated conducting sphere in Figure. The electric field is best given by:
(A) Fig (i)
(B) Fig (iii)
(C) Fig(ii)
(D) Fig(iv)
Answer:
(B) Fig (iii)
Explanation:
As given charge is + q and lines of forces in positive charge must be outwards from positive charge Question Now, as the positive charge is kept near an isolated conducting sphere, due to induction, left part of sphere gets accumulated negative charge and right part gets positive, and lines of force from right part of sphere must emerge outwards normally.
So, verifies the answer (B)
As lines of forces are not perpendicular to the surface of sphere, so options (iii) and (iv) are not true again. Hence option (C) is not correct.
Physics Chapter 1 MCQ Class 12 Question 10.
A point charge +q, is placed at a distance d from an isolated conducting plane. The field at a point P on the other side of the plane is …………..
(A) directed perpendicular to the plane and away from the plane.
(B) directed perpendicular to the plane but towards the plane.
(C) directed radially away from the point charge.
(D) directed radially towards the point charge.
Answer:
(A) directed perpendicular to the plane and away from the plane.
Explanation:
When you place a positive charge near a conducting plane, then electric field lines from positive charges will enter into the conducting plane (from the side where positive charge is kept) and emerge from opposite side of the plane. In both cases, the direction of electric field lines will always be perpendicular to the surface of the plane.
MCQ Questions Class 12 Physics Chapter 1 Question 11.
A hemisphere is uniformly charged positively. The electric field at a point on a diameter away from the centre is directed ……..
(A) perpendicular to the diameter.
(B) parallel to the diameter.
(C) at an angle tilted towards the diameter.
(D) at an angle tilted away from the diameter.
Answer:
(A) perpendicular to the diameter.
Explanation:
As the side or diameter of hemisphere is plane surface, and whole hemisphere is charged with positive charge so, the electric field line of forces emerging outward will be perpendicular to the plane surface or diameter.
Electric Charges And Fields MCQ Class 12 Question 12.
The electric flux through the surface :
(A) in Figure (iv) is the largest.
(B) in Figure (iii) is the least.
(C) in Figure (ii) is same as Figure (iii) but is smaller than Figure (iv).
(D) is the same for all the figures.
Answer:
(D) is the same for all the figures.
Explanation:
Electric flux, through the closed surface (or space) depends only on the charge enclosed inside the surface. Here, charges inside all figures are same. So, electric flux will remain same.
Physics Class 12 MCQ Chapter 1 Question 13.
Figure shows electric field lines in which an electric dipole p is placed as shown. Which of the following statements is correct?
(A) The dipole will not experience any force.
(B) The dipole will experience a force towards right.
(C) The dipole will experience a force towards left.
(D) The dipole will experience a force upwards.
Answer:
(C) The dipole will experience a force towards left.
Explanation:
We know electric field emerges radially outward from positive point charge. In the figure given above, space between field lines is increasing (or density of electric field line is decreasing). In other words, the electric force is decreasing while moving from left to right.
Thus, the force on charge – q is greater than the force on charge + q and in turn, dipole will experience a force towards left direction.
Class 12 Physics Ch 1 MCQ Question 14.
Five charges q1 q2, q3 q4, and q5 are fixed at their positions as shown in Figure. S is a Gaussian surface. The Gauss’s law is given by:
∮E.ds = q / ε0
Which of the following statements is correct?
(A) E on the LHS of the above equation will have a contribution from q1 q5and q3 while q on the RHS will have a contribution from q2 and q4 only.
(B) E on the LHS of the above equation will have a contribution from all charges, while q on the RHS will have a contribution from q2 and q4 only.
(C) E on the LHS of the above equation will have a contribution from all charges, while q on the RHS will have a contribution from qv q3 and q5 only.
(D) Both E on the LHS and q on the RHS will have contributions from q2 and q4 only.
Explanation:
As all charges are positive (or of same signs), so electric field lines on R.H.S. of Gaussian surface will be due to q2 and q4 only. On L.H.S. of Gaussian surface, the electric field lines on ‘E’ will be due to q1 q2 q3 q4 and q5So, answer (B) is verified.
MCQ Of Physics Class 12 Chapter 1 Question 15.
The Electric field at a point is –
(A) always discontinuous.
(B) discontinuous if there is a positive charge at that point.
(C) discontinuous only if there is a negative charge at that point.
(D) discontinuous if there is a charge at that point.
Answer:
(D) discontinuous if there is a charge at that point.
Explanation:
Either positive or negative charge will interact with the lines of electric field, so the electric field will become discontinuous.
MCQ Chapter 1 Physics Class 12 Question 16.
If there is only one type of charge in the universe, then –
(A) ∮E.ds ≠ 0 on any surface.
(B) ∮E.ds = Sp/q if the charge is outside the surface.
(C) ∮E.ds cannot not be defined.
(D) ∮E.ds = q/ε0 if charges of magnitude q is inside the surface.
Answer:
(D) ∮E.ds = q/ε0 if charges of magnitude q is inside the surface.
Explanation:
If a charge q is enclosed inside Gaussian surface then according to Gauss law ∮E.ds = q/ε0
Assertion And Reason Based MCQs (1 Mark each)
Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true
Class 12th Physics Chapter 1 MCQ Question 1.
Assertion (A): In a non-uniform electric field, a dipole will have translatory as well as rotatory motion.
Reason (R): In a non-uniform electric field, a dipole experiences a force as well as torque.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
When an electric dipole is placed in a uniform electric field at an angle 0 with the field, the dipole experiences a torque.
The torque produced by two parallel forces qE acting as couple = τ
τ = qE (2l sin θ)
In case of non-uniform field, force acting on both the ends of the dipole will not be equal. So, there will be a combination of couple and a net force. In this way, dipole will have both rotational as well as linear motion. So, both assertion and reason are true. Reason also explains the assertion.
Class 12 Chapter 1 Physics MCQ Question 2.
Assertion (A): The basic difference between magnetic lines of force and electric lines of force is electric lines of force are discontinuous and magnetic lines of force are continuous.
Reason (R): Magnetic lines of force exist in a magnet but no electric lines of force exists in a charged body.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
Let us consider an electric dipole. The electric lines of force exists outside only and not inside the dipole. Let us now consider a magnetic dipole. The magnetic lines of force exist outside as well as inside the dipole. So, it can be said that magnetic lines of force are continuous and electric lines of force are discontinuous. So assertion and reason both are true and reason explains the assertion too.
Chapter 1 MCQ Physics Class 12 Question 3.
Assertion (A): Electric lines of force cross each other.
Reason (R): The resultant electric field at a point is the superimposition of the electric fields at that point.
Answer:
(D) A is false and R is true
Explanation:
Electric lines of force never cross each other. If they cross each other, then at that point, we get two directions of electric field at that point, which is not possible. So, the assertion is false. The resultant electric field at a point is a vector sum of the electric fields at that point.
Class 12 Physics MCQs Chapter 1 Question 4.
Assertion (A): When bodies are charged through friction, there is transfer of charge from one body to another. No charge is created or destroyed. Reason (R): This is according to the law of conservation of electric charge.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
When two bodies are rubbed, electrons move from one body to another The body which loses electrons becomes positively charged. The body which receives the electron becomes negatively charged. So, the assertion is true. Law of conservation of electric charge states that electric charge can neither be created nor destroyed. In a closed system, the amount of charge remains same. Hence the reason is also true and properly explains the assertion.
MCQ Class 12 Physics Chapter 1 Question 5.
Assertion (A): If two spherical conductors of different radii have the same surface charge densities, then their electric field intensities will be equal.
Reason (R): Surface charge density = \(\frac {Total Charge}{Area}\)
Answer:
(B) Both A and R are true but R is NOT the correct explanation of A
Explanation:
If σ be the surface charge density of the two spheres of radius r and R, then electric fields for the two spheres are respectively:
E1 = \(\frac{\kappa 4 \pi r^{2} \sigma}{r^{2}}\) = k 4πσ
E2 = \(\frac{\kappa 4 \pi R^{2} \sigma}{R^{2}}\) = k 4πσ
So electric field intensities are equal. The assertion is true.
Surface charged density is charge per unit area = Total charge/area.
So reason is also true.
But the reason does not explain the assertion.
Class 12 Physics Chapter 1 MCQs Question 6.
Assertion (A): In a cavity in a conductor, the electric field is zero.
Reason (R): Charges in a conductor reside only at its surface.
Answer:
(A) Both A and R are true and R is the correct explanation of A
Explanation:
The charge enclosed by the Gaussian surface surrounding the cavity is zero. Hence, the electric field is also zero. So, the assertion is true. Charges in a conductor reside only at its surface. So, in cavity there is no charge. So, the reason is also true and properly explains the assertion.
Question 7.
Assertion (A): Three point charges Q1, Q2 and Q3 are shown in the figure. The flux over the Gaussian surface depends on only one charge point.
Reason (R): Electric flux depends on the all charges nearby.
Answer:
(D) A is false and R is true
Explanation:
According to Gauss’s law, electric flux depends on the enclosed charges only. Here the enclosed charges are Q1 and Q3 only. Hence the assertion is false and the reason is true.
Question 8.
Assertion (A): Using Gauss law, it is possible to find the electric field at any point.
Reason (R): Gauss law is applicable for any type of charge distribution.
Answer:
(C) A is true but R is false
Case-Based MCQs
Attempt any 4 sub-parts out of 5. Each question carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
Faraday Cage: A Faraday cage or Faraday shield is an enclosure made of conducting material. The fields within a conductor cancel out with any external fields, so the electric field within the encloser is zero. These Faraday cages act as big hollow conductors you can put things in to shield them from electrical fields. Any electrical shocks the cage receives, pass harmlessly around the outside of the cage.
Question 1.
Which of the following material can be used to make a Faraday cage ?
(A) Plastic
(B) Glass
(C) Copper
(D) Wood
Answer:
(C) Copper
Explanation:
A Faraday cage or Faraday shield is an enclosure made of a conducting material. Since copper is the only metal given in the list of options, copper is the correct answer.
Question 2.
Example of a real-world Faraday cage is :
(A) car
(B) plastic box
(C) lighting rod
(D) metal rod
Answer:
(A) car
Explanation:
Cars are example of Faraday Cages in the real world. Cars can help keep us safe from lightning. Its metal body acts as a Faraday Cage.
Question 3.
What is the electrical force inside a Faraday cage when it is struck by lightning ?
(A) The same as the lightning
(B) Half that of the lightning
(C) Zero
(D) A quarter of the lightning
Answer:
(C) Zero
Explanation:
The field within a conductor cancel out with any external fields, so the electric field within the enclosure is zero.
Question 4.
If isolated point charge +q is placed inside the Faraday cage. Its surface must have charge equal to:
(A) Zero
(B) + q
(C) – q
(D) +2q
Answer:
(C) -q
Explanation:
If a charge is placed inside an ungrounded Faraday shield without touching the walls of the internal face of the shield becomes charged with – q, and + q accumulates on the outer face of the shield. If the cage is grounded, the excess charges will be neutralized by the ground connection.
Question 5.
A point charge of 2 C is placed at centre of Faraday cage in the shape of cube with surface of 9 cm edge. The number of electric field lines passing through the cube normally will be :
(A) 1.9 x 105Nm2/C entering the surface.
(B) 1.9 x 105 Nm2/C leaving the surface.
(C) 2.0 x 105 Nm2/C leaving the surface.
(D) 2.0 x 105 Nm2/C entering the surface.
Answer:
(D) 2.0 x 105 Nm2/C entering the surface.
Explanation:
The number of electric field lines passing through the cube normally and leaving the surface = Q/ε0
Q = 2 μC = 2 x 106 C
ε0 = 8.85 x 10-12 C2/Nm2
∴ Q/ε0 = 2.2 x 105 C2/Nm2
II. Read the following text and answer the following questions on the basis of the same:
Tribo-electric series:
The tribo-electric series is a list that ranks materials according to their tendency to gain or lose electrons. The process of electron transfer as a result of two objects coming into contact with one another and then separating is called tribo-electric charging. During such an interaction, one of the two objects will always gain electrons (becoming negatively charged) and the other object will lose electrons (becoming positively charged). The relative position of the two objects on the tribo-electric series will define which object gains electrons and which object loses electrons.
In tribo-electric series, materials are ranked from high to low in terms of the tendency for the material to lose electron. If an object high up on this list (Glass, for example) is rubbed with an object low down on the list (Teflon, for example), the glass will lose electrons to the teflon. The glass will, in this case, become positively charged and the teflon will become negatively charged. Materials in the middle of the list (steel and wood, for example) are items those do not have a strong tendency to give up or accept electrons.
Question 1.
Materials in the upper position has ………….. tendency to become positively charged.
(A) low
(B) high
(C) no
(D) medium
Answer:
(B) high
Explanation:
In tribo-electric series, materials are ranked from high to low in terms of the tendency for the material to lose electron i.e. they are ranked high to low tendency of getting positively charged.
Question 2.
Name two materials which do not have a strong tendency to give up or accept electrons.
(A) Ebonite, Nylon
(B) Plastic wrap, Teflon
(C) Nylon, cat fur
(D) Steel, wood
Answer:
(D) Steel, wood
Explanation:
Materials in the middle of the list (steel and wood, for example) are items those do not have a strong tendency to give up or accept electrons.
Question 3.
If human hair is rubbed with amber, how those will be charged?
(A) Both negative
(B) Both positive
(C) Hair will be positively charged, Amber will be negatively charged.
(D) Hair will be negatively charged, Amber will be positively charged.
Answer:
(D) Hair will be negatively charged, Amber will be positively charged.
Explanation:
Since human hair is placed at the upper portion of the list, it will leave electron and will be positively charged. Since amber is placed at the lower portion of the list, it will accept the electron and will be negatively charged.
Question 4.
Tribo-electric charging is the process of electron transfer between two objects ……………
(A) By contact
(B) Without contact
(C) By any one of the above
(D) By none of the above
Answer:
(A) By contact
Explanation:
The process of electron transfer as a result of two objects coming into contact with one another and then separating is called tribo- electric charging.
Question 5.
The object which loses electron becomes …………… charged and the object gains electron becomes …………… charged.
(A) negatively, negatively
(B) positively, positively
(C) positively, negatively
(D) negatively, positively
Answer:
(C) positively, negatively
Explanation:
During tribo-electric charging, one of the two objects always gains electrons and ‘become negatively charged. The other object loses electrons and become positively charged.