NCERT Exemplar Solutions for CBSE Class 12 Physics, Chapter 2 (Part II)
NCERT Exemplar Solutions for Class 12th Physics, Chapter 2: Electrostatic Potential and Capacitance are available here. In this article, you will find solutions to multiple choice questions with multiple correct answers i.e., question number 2.7 to question number 2.13. Questions from NCERT Exemplar are frequently asked in CBSE 12th Physics board exam and other competitive exams.
NCERT Exemplar Solutions for Class 12th Physics, Chapter 2 (from question number 2.7 to 2.31) are given below:
Question 2.7: Consider a uniform electric field in the ˆz direction. The potential is a constant
(a) in all space
(b) for any x for a given z
(c) for any y for a given z
(d) on the x-y plane for a given z
Solution 2.7: (b), (c), (d)
Here, electric field is uniform and in the direction of +ve Z axis.
So, x-y plane will act as equipotential surface. In other words, potential at any point, which lies on x-y plane, x axis and y – axis will be constant.
Question 2.8: Equipotential surfaces
(a) are closer in regions of large electric fields compared to regions of lower electric fields
(b) will be more crowded near sharp edges of a conductor
(c) will be more crowded near regions of large charge densities
(d) will always be equally spaced
Solution 2.8: (a, b, c)
Equipotential surfaces are closer in regions of large electric fields because electric field intensity is inversely proportional to the separation between equipotential surfaces.
As the electric field intensities is large near sharp edges of a charged conductor or near the regions of large charge densities. Therefore, numbers of equipotential surfaces are closer to such places or in other words they are more crowded.
Question 2.9: The work done to move a charge along an equipotential from A to B
Question 2.10: In a region of constant potential
(a) the electric field is uniform
(b) the electric field is zero
(c) there can be no charge inside the region.
(d) the electric field shall necessarily change if a charge is placed outside the region.
Electric field (E) and electric potential (V) are related as E = ‒ (dV/dr).
If V is constant then it implies that electric potential at that place is zero. It may also be possible that there can be no charge inside the region.
Question 2.11: In the circuit shown in Fig. 2.4. initially key K1 is closed and key K2 is open. Then K1 is opened and K2 is closed (order is important). [Take Q1′ and Q2′ as charges on C1 and C2 and V1 and V2 as voltage respectively.]
(a) charge on C1 gets redistributed such that V1 = V2
(b) charge on C1 gets redistributed such that Q1′ = Q2′
(c) charge on C1 gets redistributed such that C1V1 + C2V2 = C1 E
(d) charge on C1 gets redistributed such that Q1′ + Q2′ = Q
Solution 2.11: (a, d)
When key K1 is dosed and K2 is open, the capacitor C1 gets charged by cell.
When K1 is opened and K2 is closed, the charge stored by capacitor C1 gets redistributed between C1 and C2 until potential of both the capacitors become same (i.e., V1 = V2).
By law of conservation of charge, total charge Q = Q1 + Q2.
Question 2.12: If a conductor has a potential V ≠ 0 and there are no charges anywhere else outside, then
(a) there must be charges on the surface or inside itself.
(b) there cannot be any charge in the body of the conductor.
(c) there must be charges only on the surface.
(d) there must be charges inside the surface.
Solution 2.12: (a), (b)
Charge always resides on the outer surface of a closed charged conductor.
Question 2.13: A parallel plate capacitor is connected to a battery as shown in Fig. 2.5. Consider two situations:
A: Key K is kept closed and plates of capacitors are moved apart using insulating handle.
B: Key K is opened and plates of capacitors are moved apart using insulating handle.
Choose the correct option(s).
a) In A: Q remains same but C changes.
(b) In B: V remains same but C changes.
(c) In A: V remains same and hence Q changes.
(d) In B: Q remains same and hence V changes.
Solution 2.13: (c), (d)
In case A, key K is kept closed and plates of capacitors are moved apart using insulating handle (i.e. distance between plates is increasing). As capacitance C = [(KεoA)/d] ⇒ C ∝ d (separation between plates) so, capacitance will decrease & amount of charge stored will also decrease (as Q = CV). Here, there will be no change in potential difference.
In case B, key K is opened and plates of capacitors are moved apart using insulating handle, by conservation of charge, charge stored by the capacitor remains same. Plates of capacitance are moving apart so capacitance will decrease and with the decreases of capacitance, potential difference V increases (as V = Q/C).