Chapter 2 Electrostatic Potential and Capacitance

NCERT Solutions Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

Class 12 Physics Chapter 2, Electrostatic Potential and Capacitance, builds on the concepts of electric charges and electric fields studied in the earlier chapter. It explains electric potential, potential difference, electrostatic potential energy, and capacitance. Students learn about the work done in moving a charge through an electric field, the energy of a system of charges, and the ability of a capacitor to store electric charge. The chapter also covers important topics such as equipotential surfaces, capacitors, dielectrics, and energy stored in a capacitor.

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance provide simple, step-by-step answers to the questions in the NCERT textbook. They help students understand important formulas, derivations, numerical problems, and concepts related to electric potential and capacitors. These solutions are useful for CBSE board exam preparation and for practising important questions for JEE and NEET.

1.0NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance: Key Concepts

Class 12 Physics Chapter 2, Electrostatic Potential and Capacitance, explains electric potential, potential energy, capacitors, and the storage of electrical energy. The main concepts covered in this chapter include:

Electrostatic Potential: Understand the work done per unit charge in bringing a positive test charge from infinity to a point in an electric field.
Potential Due to a Point Charge and Dipole: Learn how to find the electric potential due to a point charge and an electric dipole at different points. Electric potential is a scalar quantity.
Equipotential Surfaces: Understand surfaces where the electric potential is the same at every point. The electric field is always perpendicular to an equipotential surface.
Electrostatic Potential Energy: Learn about the potential energy of a system of point charges and the energy associated with their positions.
Capacitance: Understand the meaning of capacitance and study parallel plate capacitors. Learn how dielectric materials affect the capacitance of a capacitor.
Energy Stored in a Capacitor: Learn how energy is stored in a charged capacitor and understand the expression for energy density in an electric field.

2.0NCERT Solutions Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance : Detailed Solutions

SOLVED EXAMPLES

(a) Calculate the potential at a point P due to charge of $4 \times 1 0^{- 7} C$ located 9 cm away. (b) Hence obtain the work done in bringing a charge of $2 \times 1 0^{- 9} C$ from infinity to the point P . Does the answer depend on the path along which the charge is brought? Sol. (a) $V = \frac{1}{4 π ε _{0}} \frac{Q}{r}$

$= 9 \times 1 0^{9} \times \frac{4 \times 1 0 ^{- 7}}{0.09} = 4 \times 1 0^{4} V$

(b)

$W = qV = 2 \times 1 0^{- 9} \times 4 \times 1 0^{4} = 8 \times 1 0^{- 5} J$

No, work done will be path independent.

Two charges $3 \times 1 0^{- 8} C$ and $- 2 \times 1 0^{- 8} C$ are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero. Sol. Let us take the origin O at the location of the positive charge.

Electrostatic potential diagram showing two charges 15 cm apart and origin O at the positive charge.

Let P be the required point on the x -axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. If x lies between O and A,

$V = V_{1} + V_{2} = 0 \frac{1}{4 π ε _{0}} [\frac{3 \times 1 0 ^{- 8}}{x} - \frac{2 \times 1 0 ^{- 8}}{( 15 - x )}] = 0$

where x is in cm. That is,

$\Rightarrow \frac{3}{x} - \frac{2}{15 - x} = 0 45 - 3 x = 2 x 5 x = 45 \Rightarrow x = 9 cm$

Electrostatic potential diagram showing point P on the x-axis between two charges where the net potential is zero.

If x lies on the extended line OA, the required condition is $\frac{1}{4 π ε _{0}} [\frac{3 \times 1 0 ^{- 8}}{x} - \frac{2 \times 1 0 ^{- 8}}{( x - 15 )}] = 0$

$\frac{3}{x} - \frac{2}{x - 15} = 0$

Which gives $x = 45 cm$ Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge.

[Note that the formula for potential used in the calculation required choosing potential to be zero at infinity]

Figure (a) and (b) show the field lines of a positive and negative point charge respectively

Electric field line diagrams for positive and negative point charges, showing the direction of field lines around each charge.

(a) Give the signs of the potential difference $V_{P} - V_{Q}; V_{B} - V_{A}$ . (b) Give the sign of the potential energy difference of a small negative charge between the points Q and $P; A$ and B . (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A ?

Sol.

(a) As $V \propto \frac{1}{r}, V_{P} > V_{Q}$ . Thus, $(V_{P} - V_{Q})$ is positive. Also $V_{B}$ is less negative than $V_{A}$ . Thus, $V_{B} > V_{A}$ or $(V_{B} - V_{A})$ is positive. (b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, $(P.E.)_{A} > (P.E.)_{B}$ and hence sign of potential energy differences is positive. (c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative. (d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive. (e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. (a) Find the work required to put together this arrangement. (b) A charge $q_{0}$ is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?

Square arrangement of four positive and negative charges at the corners with a central point E.

Sol.

(a) Suppose, first the charge +q is brought to A, and then the charges $- q, + q$ , and -q are brought to B, C and D, respectively. The total work needed can be calculated in steps : (i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.

$i.e. W_{1} = 0$

(ii) Work needed to bring -q to B when +q is at A. $W_{2} = ($ Potential due to q charge at A $) \times$ charge at B,

$W_{2} = (\frac{q}{4 π ε _{0} d}) \times (- q) = - \frac{q ^{2}}{4 π ε _{0} d}$

(iii) Work needed to bring charge +q to C when +q is at A and -q is at B. $W_{3} =$ (potential at C due to charges at A & B) × charge at C

$W_{3} = (\frac{+ q}{4 π ε _{0} d 2} + \frac{- q}{4 π ε _{0} d}) \times q W_{3} = \frac{- q ^{2}}{4 π ε _{0} d} (1 - \frac{1}{2})$

(iv) Work needed to bring -q to D when + q at A,-q at B, and +q at C. $W_{4} =$ (potential at D due to charges at A, B and C) × charge at D.

$W_{4} W_{4} = (\frac{+ q}{4 π ε _{0} d} + \frac{- q}{4 π ε _{0} d 2} + \frac{q}{4 π ε _{0} d}) \times (- q) = \frac{- q ^{2}}{4 π ε _{0} d} (2 - \frac{1}{2})$

Add the work done in steps (i), (ii), (iii) and (iv). The total work required is,

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.
Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.

(b) The extra work necessary to bring a charge $q_{0}$ to the point E when the four charges are at A , B, C and D is $q_{0} \times$ (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to point E.

(a) Determine the electrostatic potential energy of a system consisting of two charges $7 μ C$ and $- 2 μ C$ (and with no external field) placed at (-9 cm, 0, 0) and (9 cm, 0, 0) respectively. (b) How much work is required to separate the two charges infinitely away from each other? (c) Suppose that the same system of charges is now placed in an external electric field

$E = A (\frac{1}{r ^{2}}); A = 9 \times 1 0^{5} NC^{- 1} m^{2}$

What would the electrostatic energy of the configuration be?

Sol.

(a)

$U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r} = 9 \times 1 0^{9} \times \frac{7 \times ( - 2 ) \times 1 0 ^{- 12}}{0.18} = - 0.7 J$

(b) $W = U_{2} - U_{1} = 0 - U = 0 - (- 0.7) = 0.7 J$ . (c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,

$q_{1} V (r_{1}) + q_{2} V (r_{2}) = A \frac{7 μ C}{0.09 m} + A \frac{- 2 μ C}{0.09 m}$

and the net electrostatic energy is

$q_{1} V (r_{1}) + q_{2} V (r_{2}) + \frac{q _{1} q _{2}}{4 π ε _{0} r _{12}} = A \frac{7 μ C}{0.09 m} + A \frac{- 2 μ C}{0.09 m} - 0.7 J = 70 - 20 - 0.7 = 49.3 J$

A molecule of a substance has a permanent electric dipole moment of magnitude $1 0^{- 29} Cm$ . A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude $1 0^{6} V m^{- 1}$ . The direction of the field is suddenly changed by an angle of 60°. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample. Sol. Here, dipole moment of each molecules

$= 1 0^{- 29} Cm$

As 1 mole of the substance contains $6 \times 1 0^{23}$ molecules, total dipole moment of all the molecules,

$p = 6 \times 1 0^{23} \times 1 0^{- 29} Cm = 6 \times 1 0^{- 6} Cm$

Initial potential energy,

$U_{i} = - pE cos θ = - 6 \times 1 0^{- 6} \times 1 0^{6} cos 0^{\circ} = - 6 J$

Final potential energy (when $θ = 6 0^{\circ}$ ), $U_{f} = - 6 \times 1 0^{- 6} \times 1 0^{6} cos 6 0^{\circ} = - 3 J$ Change in potential energy $= - 3 J - (- 6 J) = 3 J$ So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.

(a) A comb run through one's dry hair attracts small bits of paper. Why? What happens if the hair is wet or if it is a rainy day? (Remember, a paper does not conduct electricity.) (b) Ordinary rubber is an insulator. But special rubber tyres of aircraft are made slightly conducting. Why is this necessary? (c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. Why? (d) A bird perches on a bare high power line, and nothing happens to the bird. A man standing on the ground touches the same line and gets a fatal shock. Why?

Sol.

(a) This is because the comb gets charged by friction. The molecules in the paper gets polarised by the charged comb, resulting in a net force of attraction. If the hair is wet or if it is rainy day, friction between hair and the comb reduces. The comb does not get charged and thus it will not attract small bits of paper. (b) To enable them to conduct charge (produced by friction) to the ground; as too much of static electricity accumulated may result in spark and result in fire. (c) Reason similar to (b). (d) Current passes only when there is difference in potential.

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4) d$ , where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates? Sol. If the dielectric is now inserted, the electric field in the dielectric will be $E = \frac{E _{0}}{K}$ . The potential difference will then be

$V V ∴ σ = E_{0} (\frac{1}{4} d) + \frac{E _{0}}{K} (\frac{3}{4} d) = E_{0} d (\frac{1}{4} + \frac{3}{4 K}) = \frac{σ}{ε _{0}} \cdot d (\frac{K + 3}{4 K}) = \frac{Q}{ε _{0} A} d (\frac{K + 3}{4 K}) = \frac{Q}{A}$

The capacitance, $C = \frac{Q}{V}$

$C \Rightarrow C C = \frac{Q}{\frac{Q \cdot d}{ε _{0} A} [ \frac{K + 3}{4 K} ]} = (\frac{4 K}{K + 3}) \frac{ε _{0} A}{d} = (\frac{4 K}{K + 3}) C_{0}$

A network of four $10 μ F$ capacitors is connected to a 500 V supply, as shown in Fig. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

Sol.

Circuit diagram showing four capacitors connected in a network across a voltage supply.

(a) In the given network, $C_{1}, C_{2}$ and $C_{3}$ are connected in series. The effective capacitance C' of these three capacitors is given by

$\frac{1}{C ^{'}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$

For $C_{1} = C_{2} = C_{3} = 10 μ F, C^{'} = (10/3) μ F$ . The network has $C^{'}$ and $C_{4}$ connected in parallel. Thus, the equivalent capacitance C of the network is

$C = C^{'} + C_{4} = (\frac{10}{3} + 10) μ F = 13.3 μ F$

(b) Clearly, from the figure, the charge on each of the capacitors, $C_{1}, C_{2}$ and $C_{3}$ is the same, say Q . Let the charge on $C_{4}$ be $Q^{'}$ . Now, since the potential difference across AB is $\frac{Q}{C _{1}}$ , across BC is $\frac{Q}{C _{2}}$ , across CD is $\frac{Q}{C _{3}}$ , we have

$\frac{Q}{C _{1}} + \frac{Q}{C _{2}} + \frac{Q}{C _{3}} = 500 V$

Also, $\frac{Q ^{'}}{C _{4}} = 500 V$ This gives for the given value of the capacitance,

$Q = 500 V \times \frac{10}{3} μ F = 1.7 \times 1 0^{- 3} C and Q^{'} = 500 V \times 10 μ F = 5.0 \times 1 0^{- 3} C$

(a) A 900 pF capacitor is charged by 100 V battery [Fig. (a)]. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. (b)]. What is the electrostatic energy stored by the system?

Capacitor diagrams showing a charged capacitor before and after connection to an identical capacitor.

(a) The charge on the capacitor is

$Q = C V = 900 \times 1 0^{- 12} F \times 100 V = 9 \times 1 0^{- 8} C$

The energy stored by the capacitor is

$= \frac{1}{2} CV^{2} = \frac{1}{2} QV = \frac{1}{2} \times 9 \times 1 0^{- 8} C \times 100 V = 4.5 \times 1 0^{- 6} J$

(b) In the steady situation, the two capacitors have their positive plates at the same potential and their negative plates at the same potential. Let the common potential difference be $V^{'}$ . The charge on each capacitor is then $Q^{'} = CV^{'}$ . By charge conservation, $Q^{'} = \frac{Q}{2}$ . This implies $V^{'} = \frac{V}{2}$ . The total energy of the system is

$= 2 \times \frac{1}{2} Q^{'} V^{'} = \frac{1}{4} Q V = 2.25 \times 1 0^{- 6} J$

There is a transient period before the system settles to the situation (b). During this period, a transient current flows from the first capacitor to the second. Energy is lost during this time in the form of heat and electromagnetic radiation.

EXERCISE QUESTIONS WITH SOLUTIONS

Two charges $5 \times 1 0^{- 8} C$ and $- 3 \times 1 0^{- 8} C$ are located 16 cm apart. At what point(s) on the line joining the two charge is the electric potential zero ? Take the potential at infinity to be zero. Sol. (a) For potential at a point P (between A and B ) to be zero,

Electric potential diagram showing a point P between two opposite charges where the net potential is zero.

$\Rightarrow k [\frac{q _{1}}{x} + \frac{q _{2}}{16 - x}] = 0 \frac{q _{1}}{x} = - \frac{q _{2}}{16 - x} \frac{5 \times 1 0 ^{- 8}}{x} = \frac{3 \times 1 0 ^{- 8}}{16 - x} \frac{5}{x} = \frac{3}{16 - x} 3 x = 80 - 5 x x = 10 cm$

(b) For potential at a point $P^{'}$ (on the line joining A and B and farther from $q_{1}$ than from $q_{2}$ ) to be zero

$k [\frac{q _{1}}{y} + \frac{q _{2}}{y - 16}] = 0$

$\Rightarrow \frac{q _{1}}{y} = - \frac{q _{2}}{y - 16}$

$\frac{5 \times 1 0 ^{- 8}}{y} = \frac{- ( - 3 \times 1 0 ^{- 8} )}{y - 16}$

$\Rightarrow \frac{5}{y} = \frac{3}{y - 16} = 5 y - 80 = 3 y$

$\Rightarrow 2 y = 80 y = 40 cm$

A regular hexagon of side 10 cm has a charge $5 μ C$ at each of its vertices. Calculate the potential at the centre of the hexagon. Sol. For regular hexagon $r = a = 10 cm, q = 5 \times 1 0^{- 6} C$

Regular hexagon showing equal charges at its vertices and the centre point where electric potential is calculated.

Potential at centre

$V_{0} = \frac{6 kq}{a} = \frac{6 \times 9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 6}}{1 0 ^{- 1}} = 2.7 \times 1 0^{6} V$

Two charges $2 μ C$ and $- 2 μ C$ are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface? Sol. (a) The equatorial plane is equipotential. $(V = 0)$

Equipotential plane for two equal and opposite charges, with the equatorial plane midway between the charges.

(b) perpendicular to surface

A spherical conductor of radius 12 cm has a charge of $1.6 \times 1 0^{- 7} C$ distributed uniformly on its surface. What is the electric field? (a) Inside the sphere (b) Just outside the sphere (c) At a point 18 cm from the centre of the sphere ? Sol. Here; $q = 1.6 \times 1 0^{- 7} C, R = 12 cm = 12 \times 1 0^{- 2} m$ (a) Electric field inside the sphere = 0

(b) Electric field just outside the sphere, (on its surface),

$E = k \frac{q}{r ^{2}} = 9 \times 1 0^{9} \frac{1.6 \times 1 0 ^{- 7}}{( 12 \times 1 0 ^{- 2} ) ^{2}} = 1 0^{5} N / C$

(c)

$E = k \frac{q}{r ^{2}} \Rightarrow E = 9 \times 1 0^{9} \frac{1.6 \times 1 0 ^{- 7}}{( 18 \times 1 0 ^{- 2} ) ^{2}} = 4.4 \times 1 0^{4} N / C$

A parallel plate capacitor with air between the plates has a capacitance of $8 pF (1 pF = 1 0^{- 12} F)$ . What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6 ? Sol. Electric capacitance in air, $C = \frac{ε _{0} A}{d} = 8 pF$ New electric capacitance,

$C^{'} = \frac{ε _{r} ε _{0} A}{( d /2 )} = (2 ε_{r}) (\frac{ε _{0} A}{d}) = 2 ε_{r} C C = 2 \times 6 \times 8 pF = 96 pF$

Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination ? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ? Sol. (a) Capacitance of each of the three capacitors, $C = 9 pF$

$\frac{1}{C _{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} = \frac{3}{9} = \frac{1}{3} \Rightarrow \frac{1}{C _{eq}} = \frac{1}{3} \Rightarrow C_{eq} = 3 pF$

Therefore, total capacitance of the combination is 3pF. (b) Supply voltage, $V = 120 V$ Potential difference $(V_{1})$ across each capacitor is equal to one-third of the supply voltage.

$∴ V_{1} = \frac{V}{3} = \frac{120}{3} = 40 V$

Therefore, the potential difference across each capacitor is 40 V.

Three capacitors of capacitances 2pF, 3pF and 4pF are connected in parallel. (a) What is the total capacitance of the combination ? (b) Determine the charge on each capacitor if the combination is connected to a 100V supply. Sol. (a) $C_{1} = 2 pF, C_{2} = 3 pF$ and $C_{3} = 4 pF$ For the parallel combination of the capacitors,

$C_{e q} = C_{1} + C_{2} + C_{3} = 2 + 3 + 4 = 9 pF$

Therefore, total capacitance of the combination is 9pF. (b) Supply, voltage, $V = 100 V$ The voltage through all the three capacitor is same $= V = 100 V$

$Q_{1} = C_{1} V = 2 pF \times 100 V = (2 \times 1 0^{- 12} F) (100 V) = 2 \times 1 0^{- 10} C$

$Q_{2} = C_{2} V = 3 pF \times 100 V = (3 \times 1 0^{- 12} F) (100 V) = 3 \times 1 0^{- 10} C$

$Q_{3} = C_{3} V = 4 pF \times 100 V = (4 \times 1 0^{- 12} F) (100 V) = 4 \times 1 0^{- 10} C$

In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 1 0^{- 3} m^{2}$ and the distance between the plates is 3 mm . Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? Sol. Capacitance of parallel plate capacitor,

$C C = \frac{ε _{0} A}{d}, we get = \frac{( 8.854 \times 1 0 ^{- 12} ) ( 6 \times 1 0 ^{- 3} )}{3 \times 1 0 ^{- 3}} = 17.708 \times 1 0^{- 12} = 18 pF$

Using

$q = CV q = 18 \times 1 0^{- 12} \times 100 = 1.8 \times 1 0^{- 9} C$

Explain what would happen if in the capacitor given in Question.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. Sol. (a) (i) $C^{'} = KC = 6 \times 18 = 108 pF$ (ii) $q = C^{'} V = 108 \times 1 0^{- 12} \times 100$ $= 1.08 \times 1 0^{- 8} C$ (b) q remains $1.8 \times 1 0^{- 9} C$ ; Capacitance $= 108 pF$ $∴ V = \frac{q}{C} = \frac{1.8 \times 1 0 ^{- 9}}{108 \times 1 0 ^{- 12}} = 16.6 V$ .
A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? Sol. $C = 12 pF = 12 \times 1 0^{- 12} F, V = 50 V$ Electrostatic energy stored in the capacitor

$U = \frac{1}{2} CV^{2} = \frac{1}{2} \times 12 \times 1 0^{- 12} \times (50)^{2} J = 1.5 \times 1 0^{- 8} J$

Therefore, the electrostatic energy stored in the capacitor is $1.5 \times 1 0^{- 8} J$ .

A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ? Sol. Initial potential energy, $U_{i} = \frac{1}{2} C_{1} V_{1}^{2}$

$= \frac{1}{2} (6 \times 1 0^{- 10}) (200)^{2} = 12 \times 1 0^{- 6} J$

Common potential,

$V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}} = \frac{( 6 \times 1 0 ^{- 10} ) \times 200 + 0}{6 \times 1 0 ^{- 10} + 6 \times 1 0 ^{- 10}} V = 100 V U_{f} = \frac{1}{2} C_{1} V^{2} + \frac{1}{2} C_{2} V^{2} = \frac{1}{2} (C_{1} + C_{2}) V^{2} = \frac{1}{2} (6 \times 1 0^{- 10} + 6 \times 1 0^{- 10}) (100)^{2} J = 6 \times 1 0^{- 6} J$

Loss in energy $= U_{i} - U_{f}$

$= 12 \times 1 0^{- 6} J - 6 \times 1 0^{- 6} J = 6 \times 1 0^{- 6} J$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics chapter-wise, including exercise answers, important formulas, and step-by-step solutions to strengthen understanding and support effective practice.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 2

Clear Understanding of Scalar Potential: The solutions clearly explain electric potential as a scalar quantity and help students understand how it differs from the electric field.
Step-by-Step Derivations: Important derivations, such as the potential due to an electric dipole and the energy stored in a capacitor, are explained step by step for easy understanding and revision.
Understanding Dielectrics: The solutions explain how dielectric materials affect the capacitance, electric field, and potential of a capacitor.
Capacitor Circuit Problems: Questions based on capacitors connected in series and parallel are explained with clear steps, making it easier to understand and solve circuit problems.
Exam-Ready Solutions: Answers include the important formulas, steps, units, and explanations needed to write clear solutions in CBSE board exams.

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NCERT Solutions Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance

1.0NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 2 - Electrostatic Potential and Capacitance : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 2

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