NCERT Exemplar solutions for Class 12 Physics, Chapter 2 Electrostatic Potential and Capacitance are available here. In this article, you will get solutions to long answer type questions (i.e., question number 2.24 to question number 2.33). These questions are quite completed. Solutions to MCQ I, MCQ II, VSA & SA are already available in **Part I**, **Part II**, **Part III** and **Part IV**. These questions are important for CBSE Class 12 Physics board exam & other competitive exams like NEET, JEE Main, WBJEE etc.

NCERT Exemplar Solutions for Class 12th Physics, Chapter 2 (from question number 2.24 to 2.33) are given below:

**Question 2.24:** Find the equation of the equipotentials for an infinite cylinder of radius *r*_{0}, carrying charge of linear density λ.

**Solution 2.24:**

To find the potential at distance *r* from the line consider the electric field. We note that from symmetry the field lines must be radially outward. Draw a cylindrical Gaussian surface of radius *r* and length *l*. Then,

**Question 2.25:** Two point charges of magnitude +*q* and ‒*q* are placed at (-*d*/2, 0, 0) and (*d*/2, 0, 0), respectively. Find the equation of the equipoential surface where the potential is zero.

**Solution 2.25:**

The potential at the point P is given by

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

**Question 2.26:** A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (*U*) as ε = α*U* where α = 2V^{–1}. A similar capacitor with no dielectric is charged to U_{0} = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

**Solution 2.26:**

Let the final voltage be U: If C is the capacitance of the capacitor without the dielectric, then the charge on the capacitor is

*Q*_{1} = *CU*

The capacitor with the dielectric has a capacitance eC. Hence the charge on the capacitor is

*Q*_{2} = *eU* = *αCU*^{2}

The initial charge on the capacitor that was charged is

*Q _{o}* =

*CU*

_{o}From the conservation of charges,

*Q*_{o} = *Q*_{1} + *Q*_{2}

Or, CU_{o} = CU + *α*CU^{2}

⇒ α U^{2} + U ‒ u_{o} = 0

**Question 2.27:** A capacitor is made of two circular plates of radius R each, separated by a distance *d* << *R*. The capacitor is connected to a constant voltage. A thin conducting disc of radius *r *<< *R* and thickness *t* << *r* is placed at a centre of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is *m*.

**Solution 2.27:**

When the disc is in touch with the bottom plate, the entire plate is a equipotential. A change *q*’ is transferred to the disc.

The electric field on the disc is = V/d

**Question 2.28:** (*a*) In a quark model of elementary particles, a neutron is made ofone up quarks [charge (2/3) *e*] and two down quarks [charges – (1/3) *e*]. Assume that they have a triangle configuration with side length of the order of 10–15 m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.

(*b*) Repeat above exercise for a proton which is made of two up and one down quark.

**Solution 2.28:**

**Question 2.29:** 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?

**Solution 2.29:**

Before contact,

Q_{1} = (s.4p*R*^{2})

Q_{1} = s.4p (2*R*^{2}) = (s.4p*R*^{2}) = 4*Q*_{1}

After contact:

Q_{1}’ +Q_{2}’ = Q_{1} + Q_{2} =5Q = 5(s.4p*R*^{2})

They will be at equal potentials:

Q_{1}’/R = Q_{2}’/2R

∴ Q_{2}’ = 2Q_{1}’

∴ 3Q_{1}’ = 5(s.4p*R*^{2}) and Q_{2} = 10/3(s.4p*R*^{2})

∴ s_{1} = (5/3)s and s_{2} = (5/6) s.

**Question 2.30:** In the circuit shown in Fig. 2.7, initially K_{1} is closed and K_{2} is open. What are the charges on each capacitors. Then K_{1} was opened and K_{2} was closed (order is important), What will be the charge on each capacitor now? [C = 1μF]

**Solution 2.30:**

Initially, V ∝ (1/C) and V_{1} + V_{2} =E

⇒ V_{1} = 3 V and V_{2} = 6V

∴ Q_{1} = C_{1}V = 6C × 3 = 18 *μC *

Q_{2} = 9* μC* and Q_{3} = 0.

Later: Q_{2} = Q_{2}’ + Q_{3}

With C_{2}V + C_{3}V = Q_{2}

⇒ V = Q_{2}/(C_{2} + C_{3}) = (3/2) V

Q_{2}’ = (9/2) μC and Q_{3}’ = (9/2) μC.

**Question 2.31:** Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

**Solution 2.31:**

**Question 2.32:** Two charges *q*_{1} and *q*_{2 }are placed at (0, 0, *d*) and (0, 0, –*d*) respectively. Find locus of points where the potential a zero.

**Solution 2.32:**

The situation given in question is shown in the figure given below

**Question 2.33:** Two charges –q each are separated by distance 2d. A third charge + *q* is kept at mid point O. Find potential energy of + q as a function of small distance x from O due to – q charges. Sketch P.E. v/s *x* and convince yourself that the charge at O is in an unstable equilibrium.

**Solution 2.33:**

**NCERT Solutions for Class 12 Physics**