Get NCERT Exemplar solutions for Class 12 Physics, Chapter 2 Electrostatic Potential and Capacitance. Here, you will find solutions to short answer type questions (i.e., question number 2.14 to question number 2.18). Solutions of question number 2.19 to 2.23). Solutions to MCQ I, MCQ II and VSA are already available in **Part I**, **Part II** and **Part III**. Solutions to LA will be available in further part. These questions are important for CBSE Class 12 Physics board exam & other competitive exams like JEE Main, WBJEE etc.

*NCERT Exemplar Solutions for Class 12th Physics, Chapter 2 (from question number 2.19 to 2.23) are given below:*

**Question 2.19:** Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

**Solution 2.19:**

Equation, *E* = ‒d*V*/d*r* shows that electric potential decreases along the direction of electric field.

Suppose this were not true. The potential just inside the surface would be different from that at the surface resulting in a potential gradient. This would mean that there are field lines pointing inwards or outwards from the surface. These lines cannot at the other end be again on the surface, since the surface is equipotential. Thus, this is possible only if the other ends of the lines are at charges inside, contradicting the premise. Hence, the entire volume inside must be at the same potential.

**Question 2.20:** A capacitor has some dielectric between its plates, and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is removed. State whether the capacitance, the energy stored in it, electric field, charge stored and the voltage will increase, decrease or remain constant.

**Solution 2.20:**

C will decrease

Energy stored = ½(*CV*^{2}) and hence will increase.

Electric field will increase.

Charge stored will remain the same.

*V* will increase.

**NCERT Solutions for Class 12 Physics, Chapter 2 Electrostatic Potential and Capacitance**

**Question 2.21:** Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

**Solution 2.21:**

If we analyse the equation, E = ‒d*V*/d*r*,then we observe that electric potential decreases along the direction of electric field.

Now, consider any path from the charged conductor to the uncharged conductor along the electric field. The potential will continually decrease along this path. A second path from the uncharged conductor to infinity will again continually lower the potential further. Therefore, we can say that the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

**Question 2.22:** Calculate potential energy of a point charge –*q* placed along the axis due to a charge +*Q* uniformly distributed along a ring of radius *R*. Sketch P.E. as a function of axial distance z from the centre of the ring. Looking at graph, can you see what would happen if -q is displaced slightly from the centre of the ring (along the axis)?

**Solution 2.22:**

In the given question, charge *Q* is distributed uniformly along the ring.

**Question 2.23:** Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.

**Solution 2.23:**

The derivation is similar to the derivation given in previous question.

Final answer must be

**NCERT Solutions for Class 12 Physics**