Class 12 Physics Chapter 2 Important Questions (2026-27)
Important Questions for Class 12 Physics Chapter 2 – Electrostatic Potential and Capacitance
Looking for Class 12 Physics Chapter 2 important questions? This page covers Electrostatic Potential and Capacitance with exam-focused questions and step-by-step answers, arranged by marks: 1 mark, 2–3 marks and 5 marks. Written for the CBSE 2026–27 syllabus, it focuses on electric potential, capacitors and dielectrics. Attempt them after finishing the NCERT chapter.
Very Short Answer Questions (1 Mark)
Electric Potential and Work Done
Question 1
A 500 µC charge is at the centre of a square of side 10 cm. Find the work done in moving a charge of 10 µC between two diagonally opposite points on the square.
Question 2
What is the electrostatic potential due to an electric dipole at an equatorial point?
Question 3
What is the work done in moving a test charge \(q\) through a distance of 1 cm along the equatorial axis of an electric dipole?
Question 4
Define the term 'potential energy' of charge \(q\) at a distance \(V\) in an external electric field.
Question 5
A point charge \(Q\) is placed at point \(O\). Is the potential difference \(V_A - V_B\) positive, negative or zero, if \(Q\) is (i) positive (ii) negative?
(i) If \(Q\) is positive, \(V_A - V_B\) is positive.
(ii) If \(Q\) is negative, \(V_A - V_B\) is negative.
Potential Inside a Hollow Metal Sphere
Question 6
A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere?
Question 7
A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 5 V. What is the potential at the centre of the sphere?
Question 8
Why is electrostatic potential constant throughout the volume of the conductor and has the same value (as inside) on its surface?
Question 9
Why must the electrostatic potential inside a hollow charged conductor be the same at every point?
Dielectrics and Conductors
Question 10
Distinguish between a dielectric and a conductor.
| Dielectric | Conductor |
|---|---|
| Dielectrics are insulating materials which transmit electric effects without conducting. | Conductors are substances which can carry or conduct electric charge from one place to another. |
Equipotential Surfaces
Question 11
What is the geometrical shape of equipotential surfaces due to a single isolated charge?
Question 12
Two charges 2 µC and –2 µC are placed at points A and B 5 cm apart. Depict an equipotential surface of the system.
Question 13
What is the amount of work done in moving a point charge around a circular arc of radius \(r\) at the centre of which another point charge is located?
Question 14
Two charges 4 µC and –4 µC are placed at points A and B 3 cm apart. Depict an equipotential surface of the system.
Question 15
“For any charge configuration, equipotential surface through a point is normal to the electric field.” Justify.
\(W = FS\cos\theta \therefore \cos\theta = 0\) or \(\theta = 90^\circ\).
Capacitors and Dielectrics
Question 16
Two equal balls having equal positive charge \(q\) coulombs are suspended by two insulating strings of equal length. What would be the effect on the force when a plastic sheet is inserted between the two?
Question 17
The given graph shows variation of charge \(q\) versus potential difference \(V\) for two capacitors \(C_1\) and \(C_2\). Both capacitors have the same plate separation but the plate area of \(C_2\) is greater than that of \(C_1\). Which line (A or B) corresponds to \(C_1\) and why?
Reason: Since the slope \(\left(\frac{q}{V}\right)\) of B is less than that of A, and \(C = \frac{q}{V}\), line B has the smaller capacitance.
Question 18
A charge \(q\) is moved from a point A above a dipole of dipole moment \(p\) to a point B below the dipole in the equatorial plane without acceleration. Find the work done in the process.
\(W = qV_{AB} = q \times 0 = 0\), since the potential remains constant (zero) along the equatorial plane.
Short Answer Questions (2–3 Marks)
Electric Potential Due to a Dipole
Question 19
Derive the expression for the electric potential at any point along the axial line of an electric dipole.
Electric potential at point \(P\) due to the dipole is
\(V = V_1 + V_2\)
\(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r - a} + \dfrac{1}{4\pi\varepsilon_0}\dfrac{(-q)}{r + a}\)
\(V = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{r - a} - \dfrac{1}{r + a}\right] = \dfrac{q}{4\pi\varepsilon_0}\dfrac{2a}{r^2 - a^2}\)
Since \(p = q \times 2a\):
\(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2 - a^2}\)
For \(r \gg a\): \(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}\)
Question 20
Derive an expression for the potential energy of an electric dipole of dipole moment \(\vec{p}\) in the electric field \(\vec{E}\).
The dipole experiences no net force in a uniform electric field but experiences a torque \(\tau\) given by \(\tau = pE\sin\theta\).
This torque tends to rotate the dipole. Work must be done to rotate the dipole against this torque.
Work done in rotating the dipole from angle \(\theta_1\) to \(\theta_2\):
\(W = pE(\cos\theta_1 - \cos\theta_2)\)
Taking \(\theta_1 = 90^\circ\) (reference) and \(\theta_2 = \theta\):
\(U = -pE\cos\theta = -\vec{p}\cdot\vec{E}\)
Question 21
An electric dipole of length 4 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of \(4\sqrt{3}\) Nm. Calculate the potential energy of the dipole, if it has charge ±8 nC.
\(\tau = pE\sin\theta \Rightarrow pE = \dfrac{\tau}{\sin 60^\circ} = \dfrac{4\sqrt{3}}{\sqrt{3}/2} = 8\ \text{Nm}\)
\(U = -pE\cos\theta = -8 \times \cos 60^\circ = -8 \times 0.5 =\) \(-4\ \text{J}\)
Question 22
An electric dipole of length 2 cm, when placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of \(8\sqrt{3}\) Nm. Calculate the potential energy of the dipole, if it has a charge of ±4 nC.
\(pE = \dfrac{\tau}{\sin 60^\circ} = \dfrac{8\sqrt{3}}{\sqrt{3}/2} = 16\ \text{Nm}\)
\(U = -pE\cos\theta = -16 \times 0.5 =\) \(-8\ \text{J}\)
Question 23
An electric dipole of length 1 cm, which is placed with its axis making an angle of 60° with a uniform electric field, experiences a torque of \(6\sqrt{3}\) Nm. Calculate the potential energy of the dipole, if it has a charge of ±2 nC.
\(pE = \dfrac{\tau}{\sin 60^\circ} = \dfrac{6\sqrt{3}}{\sqrt{3}/2} = 12\ \text{Nm}\)
\(U = -pE\cos\theta = -12 \times 0.5 =\) \(-6\ \text{J}\)
Potential Energy of Systems of Charges
Question 24
Two point charges \(q_1 = 10 \times 10^{-8}\ \text{C}\) and \(q_2 = -2 \times 10^{-8}\ \text{C}\) are separated by a distance of 60 cm in air. (i) Find at what distance from the first charge \(q_1\) would the electric potential be zero. (ii) Also calculate the electrostatic potential energy of the system.
(i) Let \(AP = x\). Potential at P is zero:
\(\dfrac{kq_1}{x} + \dfrac{kq_2}{0.6 - x} = 0 \Rightarrow \dfrac{10}{x} = \dfrac{2}{0.6 - x}\)
\(10(0.6 - x) = 2x \Rightarrow 6 - 10x = 2x \Rightarrow 12x = 6 \Rightarrow x = 0.5\ \text{m}\)
Distance from first charge = 0.5 m = 50 cm
(ii) Electrostatic potential energy:
\(U = \dfrac{kq_1q_2}{r} = \dfrac{9 \times 10^9 \times 10 \times 10^{-8} \times (-2 \times 10^{-8})}{0.6}\)
\(U = \dfrac{-1.8 \times 10^{-5}}{0.6} =\) \(-3 \times 10^{-6}\ \text{J}\)
Question 25
Two point charges \(20 \times 10^{-6}\ \text{C}\) and \(-4 \times 10^{-6}\ \text{C}\) are separated by a distance of 50 cm in air. (i) Find the point on the line joining the charges where the electric potential is zero. (ii) Also find the electrostatic potential energy of the system.
(i) Let \(AP = x\), then \(PB = 0.5 - x\). Potential at P is zero:
\(\dfrac{kq_1}{x} + \dfrac{kq_2}{0.5 - x} = 0 \Rightarrow \dfrac{20}{x} = \dfrac{4}{0.5 - x}\)
\(20(0.5 - x) = 4x \Rightarrow 10 - 20x = 4x \Rightarrow 24x = 10 \Rightarrow x = \dfrac{5}{12}\ \text{m} \approx 0.417\ \text{m}\)
Point is at 0.417 m from the 20 µC charge (between the charges).
(ii) \(U = \dfrac{kq_1q_2}{r} = \dfrac{9 \times 10^9 \times 20 \times 10^{-6} \times (-4 \times 10^{-6})}{0.5} =\) \(-1.44\ \text{J}\)
Question 26
Two point charges \(4Q\) and \(Q\) are separated by 1 m in air. At what point on the line joining the charges is the electric field intensity zero? Also calculate the electrostatic potential energy of the system of charges, taking \(Q = 2 \times 10^{-7}\ \text{C}\).
\(\dfrac{k(4Q)}{x^2} = \dfrac{kQ}{(1 - x)^2} \Rightarrow \dfrac{4}{x^2} = \dfrac{1}{(1 - x)^2}\)
\(\dfrac{2}{x} = \dfrac{1}{1 - x} \Rightarrow 2 - 2x = x \Rightarrow 3x = 2 \Rightarrow x = \dfrac{2}{3}\ \text{m}\)
Point is at 2/3 m from the 4Q charge.
\(U = \dfrac{k(4Q)(Q)}{1} = 9 \times 10^9 \times 4 \times (2 \times 10^{-7})^2 =\) \(1.44 \times 10^{-3}\ \text{J}\)
Question 27
Calculate the work done to dissociate the system of three charges placed on the vertices of a triangle: \(q_1 = 1\ \mu\text{C}\), \(q_2 = 1\ \mu\text{C}\), \(q_3 = -4\ \mu\text{C}\) on an equilateral triangle of side 10 cm.
\(U_i = \dfrac{k}{r}(q_1q_2 + q_2q_3 + q_3q_1)\)
\(= \dfrac{9 \times 10^9}{0.1}[(1)(1) + (1)(-4) + (-4)(1)] \times 10^{-12}\)
\(= \dfrac{9 \times 10^9}{0.1}(1 - 4 - 4) \times 10^{-12} = \dfrac{9 \times 10^9}{0.1}(-7) \times 10^{-12}\)
\(U_i = -0.63\ \text{J}\)
Final energy \(U_f = 0\). Work done to dissociate = \(U_f - U_i =\) \(0.63\ \text{J}\)
Question 28
Two point charges \(q_1\) and \(q_2\) are located at \(\vec{r}_1\) and \(\vec{r}_2\) respectively in an external electric field \(\vec{E}\). Obtain the expression for the total work done in assembling this configuration.
\(W_1 = q_1V(\vec{r}_1)\)
Work done in bringing \(q_2\) to position \(\vec{r}_2\) against the external field and the field due to \(q_1\):
\(W_2 = q_2V(\vec{r}_2) + \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}\)
Total work done:
\(W = q_1V(\vec{r}_1) + q_2V(\vec{r}_2) + \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}\)
Equipotential Surfaces
Question 29
(i) Can two equipotential surfaces intersect each other? Give reasons. (ii) Two charges \(-q\) and \(+q\) are located at points A (0, 0, –a) and B (0, 0, +a) respectively. How much work is done in moving a test charge from point P (7, 0, 0) to Q (–3, 0, 0)?
(i) No. If they intersect, there would be two different directions of electric field at that point, which is not possible. Also, at the same point of intersection, there would be two values of potential — which is impossible.
(ii) Both points P and Q lie on the equatorial line of the dipole, where \(V = 0\) everywhere. Therefore, work done will be zero. The force on any charge is perpendicular to the equatorial line, so work done is zero.
Question 30
Draw 3 equipotential surfaces corresponding to a field that uniformly increases in magnitude but remains constant along the Z-direction. How are these surfaces different from those of a constant electric field along the Z-direction?
Question 31
Two uniformly large parallel thin plates having charge densities \(+\sigma\) and \(-\sigma\) are kept in the X-Z plane at a distance \(d\) apart. Sketch an equipotential surface due to the electric field between the plates. If a particle of mass \(m\) and charge \(q'\) remains stationary between the plates, what is the magnitude and direction of this field?
For a particle of charge \(-q\) at rest between the plates:
(i) weight \(mg\) acts vertically downward
(ii) electric force \(qE\) acts vertically upward
So \(mg = qE \Rightarrow E = \dfrac{mg}{q}\), directed vertically downward (along the negative Y-axis).
Question 32
Two small identical electrical dipoles AB and CD, each of dipole moment \(p\), are kept at an angle of 120° as shown. What is the resultant dipole moment of this combination? If this system is subjected to an electric field \(\vec{E}\) directed along the +X direction, what will be the magnitude and direction of the torque acting on it?
\(p_R = \sqrt{p^2 + p^2 + 2p^2\cos 120^\circ} = \sqrt{2p^2 - p^2} = p\)
The resultant dipole moment makes an angle of 60° with each dipole and 30° with the x-axis.
Torque \(\tau = p_R E \sin 30^\circ = pE \times 0.5 = \dfrac{pE}{2}\)
Direction of torque: perpendicular to both \(\vec{p}_R\) and \(\vec{E}\), i.e., along the z-axis (using right-hand rule).
Question 33
Define an equipotential surface. Draw equipotential surfaces: (i) in the case of a single point charge and (ii) in a constant electric field in the Z-direction. Why are the equipotential surfaces about a single charge not equidistant? (iii) Can electric field exist tangential to an equipotential surface? Give reason.
(i) For a single point charge: concentric spheres centred on the charge.
(ii) In a constant electric field in the Z-direction: parallel planes perpendicular to the Z-axis.
Equipotential surfaces are not equidistant because \(V \propto \frac{1}{r}\), so the same potential difference requires larger spacing at larger \(r\).
(iii) No. If the field lines were tangential, work would be done in moving a charge on the surface, which goes against the definition of an equipotential surface.
Capacitors and Combinations
Question 34
Figure shows two identical capacitors \(C_1\) and \(C_2\), each of 2 µF capacitance, connected to a battery of 5 V. Initially switch S is left open and dielectric slabs of dielectric constant \(K = 5\) are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?
\(q_1 = C_1V = 5CV = 5 \times 2 \times 5 = 50\ \mu\text{C}\), \(q_2 = 50\ \mu\text{C}\)
Charge on each capacitor becomes 5 times.
(ii) Potential difference across \(C_1\) is still 5 V (since battery is connected). Across \(C_2\), the p.d. becomes \(V' = \dfrac{q}{5C} = \dfrac{50}{10} = 5\ \text{V}\).
So the p.d. remains 5 V for both capacitors.
Question 35
Figure shows two identical capacitors \(C_1\) and \(C_2\), each of 1.5 µF capacitance, connected to a battery of 2 V. Initially switch S is closed. After sometime S is left open and dielectric slabs of dielectric constant \(K = 2\) are inserted to fill completely the space between the plates of the two capacitors. How will the (i) charge and (ii) potential difference between the plates of the capacitors be affected after the slabs are inserted?
\(q_1 = 2CV = 2 \times 1.5 \times 2 = 6\ \mu\text{C}\), \(q_2 = 6\ \mu\text{C}\)
Charge on each capacitor becomes twice.
(ii) Potential difference across \(C_1\) is still 2 V. Across \(C_2\), the p.d. becomes \(V' = \dfrac{q}{2C} = \dfrac{6}{3} = 2\ \text{V}\).
So the p.d. remains 2 V for both capacitors.
Question 36
Net capacitance of three identical capacitors in series is 1 pF. What will be their net capacitance if connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the same source.
In parallel: \(C_p = 3C = 9\ \text{pF}\)
Energy ratio: \(\dfrac{U_p}{U_s} = \dfrac{\frac{1}{2}C_pV^2}{\frac{1}{2}C_sV^2} = \dfrac{C_p}{C_s} = \dfrac{9}{1} =\) \(9:1\)
Question 37
Net capacitance of three identical capacitors in series is 3 pF. What will be their net capacitance if connected in parallel? Find the ratio of energy stored in the two configurations if they are both connected to the same source.
In parallel: \(C_p = 3C = 27\ \text{pF}\)
Energy ratio: \(\dfrac{U_p}{U_s} = \dfrac{C_p}{C_s} = \dfrac{27}{3} =\) \(9:1\)
Question 38
Determine the potential difference across the plates of the capacitor \(C_1\) of the network shown in the figure. [Assume \(E_2 > E_1\)]
Charge \(q = C_s(E_2 - E_1)\)
Potential difference across \(C_1\): \(V_1 = \dfrac{q}{C_1} = \dfrac{C_2(E_2 - E_1)}{C_1 + C_2}\)
Question 39
A network of four capacitors, each of capacitance 15 µF, is connected across a battery of 100 V. Find the net capacitance and the charge on the capacitor \(C_4\).
\(C_s\) and \(C_4\) are in parallel: \(C_{eq} = 5 + 15 = 20\ \mu\text{F}\).
Charge on \(C_4\): \(Q_4 = C_4V = 15 \times 10^{-6} \times 100 =\) \(1.5 \times 10^{-3}\ \text{C} = 1500\ \mu\text{C}\)
Question 40
A network of four capacitors, each of capacitance 30 pF, is connected across a battery of 60 V. Find the net capacitance and the energy stored in each capacitor.
Charge on \(C_4\): \(Q_4 = C_4V = 30 \times 10^{-12} \times 60 = 1.8 \times 10^{-9}\ \text{C}\)
Energy in \(C_4\): \(U_4 = \dfrac{1}{2}C_4V^2 = \dfrac{1}{2} \times 30 \times 10^{-12} \times 3600 = 5.4 \times 10^{-8}\ \text{J}\)
Charge on each of \(C_1, C_2, C_3\): \(Q_s = C_sV = 10 \times 10^{-12} \times 60 = 6 \times 10^{-10}\ \text{C}\)
Energy in each: \(U_s = \dfrac{Q_s^2}{2C} = \dfrac{(6 \times 10^{-10})^2}{2 \times 30 \times 10^{-12}} = 6 \times 10^{-9}\ \text{J}\)
Question 41
(a) A charge +Q is placed on a large spherical conducting shell of radius R. Another small conducting sphere of radius r carrying charge q is introduced inside the large shell and is placed at its centre. Find the potential difference between two points, one lying on the sphere and the other on the shell. (b) How would the charge between the two flow if they are connected by a conducting wire? Name the device which works on this fact.
\(V_r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r} + \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\)
Potential at the outer shell:
\(V_R = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R} + \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\)
Potential difference: \(V_r - V_R = \dfrac{1}{4\pi\varepsilon_0}q\left(\dfrac{1}{r} - \dfrac{1}{R}\right)\)
(b) If connected by a wire, charge will flow from the smaller sphere (higher potential, since \(r < R\)) to the large shell. The Van de Graaff generator works on this principle.
Question 42
Two point charges \(q\) and \(-2q\) are kept at distance \(d\) apart. Find the location of the point relative to charge \(q\) at which potential due to this system of charges is zero.
\(\dfrac{kq}{x} + \dfrac{k(-2q)}{d + x} = 0 \Rightarrow \dfrac{1}{x} = \dfrac{2}{d + x}\)
\(d + x = 2x \Rightarrow x = d\)
Point is at distance \(d\) to the left of charge \(q\) (away from \(-2q\)).
Question 43
Two closely spaced equipotential surfaces A and B with potentials \(V\) and \(V + \delta V\) (where \(\delta V\) is the change in \(V\)) are kept \(\delta l\) distance apart. Deduce the relation between the electric field and the potential gradient between them. Write two important conclusions concerning the relation between the electric field and electric potential.
\(W = -\delta V = E \times \delta l\)
\(E = -\dfrac{\delta V}{\delta l}\)
Two conclusions:
(i) Electric field is in the direction in which the potential decreases steepest.
(ii) The magnitude of the electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface.
Question 44
Why does current in a steady state not flow in a capacitor connected across a battery? However momentary current does flow during charging or discharging of the capacitor. Explain.
During charging and discharging, the displacement current and hence the conduction current is non-zero, as the electric field \(\vec{E}\) between the plates changes with time.
Dielectrics in Capacitors
Question 45
A parallel plate capacitor is charged by a battery. After some time the battery is disconnected and a dielectric slab of dielectric constant K is inserted between the plates. How would (i) the capacitance, (ii) the electric field between the plates, and (iii) the energy stored in the capacitor be affected? Justify your answer.
After disconnecting the battery, charge \(Q\) remains constant. New values: \(C' = KC\), \(V' = \dfrac{Q}{KC} = \dfrac{V}{K}\), \(E' = \dfrac{V'}{d} = \dfrac{E}{K}\).
(i) New capacitance is \(K\) times its original value (\(C' = KC\)).
(ii) Electric field becomes \(\dfrac{1}{K}\) times its original value (\(E' = E/K\)).
(iii) Energy stored: \(U' = \dfrac{Q^2}{2C'} = \dfrac{Q^2}{2KC} = \dfrac{U}{K}\), so the energy becomes \(\dfrac{1}{K}\) times its original value.
Question 46
A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance, (ii) electric field between the plates, and (iii) the energy stored in the capacitor be affected? Justify your answer in each case.
Question 47
A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. The battery used to charge it remains connected. A dielectric slab of thickness d and dielectric constant k is now placed between the plates. What change, if any, will take place in: (i) charge on plates? (ii) electric field intensity between the plates? (iii) capacitance of the capacitor? Justify your answer in each case.
(i) New charge \(Q' = C'V' = kCV = kQ\). Charge becomes \(k\) times.
(ii) Electric field \(E' = \dfrac{V'}{d} = \dfrac{V}{d} = E\). Electric field remains unchanged.
(iii) New capacitance \(C' = kC = \dfrac{k\varepsilon_0 A}{d}\). Capacitance becomes \(k\) times.
Question 48
A parallel plate capacitor is charged to a potential difference V by a d.c. source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change: (i) electric field between the plates, (ii) capacitance, and (iii) energy stored in the capacitor.
If distance \(d\) is doubled: \(C' = \dfrac{\varepsilon_0 A}{2d} = \dfrac{C}{2}\)
(i) \(E' = \dfrac{\sigma}{\varepsilon_0} = \dfrac{q}{A\varepsilon_0}\). Since \(q\) is unchanged, \(E\) remains unchanged.
(ii) Capacitance becomes half (\(C' = C/2\)).
(iii) Energy \(U = \dfrac{q^2}{2C}\). Since \(C\) halves, \(U' = \dfrac{q^2}{2(C/2)} = 2U\). Energy becomes twice.
Question 49
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has thickness d/2, where d is the separation between the plates. Find the expression for its capacitance when the slab is inserted between the plates of the capacitor.
\(C_1 = \dfrac{K\varepsilon_0 A}{d/2} = \dfrac{2K\varepsilon_0 A}{d}\), \(C_2 = \dfrac{\varepsilon_0 A}{d/2} = \dfrac{2\varepsilon_0 A}{d}\)
\(\dfrac{1}{C} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{d}{2K\varepsilon_0 A} + \dfrac{d}{2\varepsilon_0 A} = \dfrac{d}{2\varepsilon_0 A}\left(\dfrac{1}{K} + 1\right)\)
\(C = \dfrac{2\varepsilon_0 A}{d} \cdot \dfrac{K}{K + 1}\)
Capacitance of Parallel Plate Capacitor
Question 50
Derive the expression for the capacitance of a parallel plate capacitor having plate area A and plate separation d.
Electric field between the plates due to each plate \(= \dfrac{\sigma}{2\varepsilon_0}\). The fields add up:
\(E = \dfrac{\sigma}{2\varepsilon_0} + \dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{\varepsilon_0 A}\)
Potential difference: \(V = Ed = \dfrac{Qd}{\varepsilon_0 A}\)
Capacitance: \(C = \dfrac{Q}{V} =\) \(\dfrac{\varepsilon_0 A}{d}\)
Long Answer Questions (5 Marks)
Energy Stored in a Capacitor
Question 51
Deduce the expression for the electrostatic energy stored in a capacitor of capacitance C and having charge Q. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant K?
\(dW = V\,dq = \dfrac{q}{C}\,dq\)
Total work done in giving charge \(Q\):
\(W = \int_0^Q \dfrac{q}{C}\,dq = \dfrac{Q^2}{2C}\)
This work is stored as potential energy: \(U = \dfrac{Q^2}{2C} = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV\)
(i) With dielectric of constant K (battery disconnected), \(C' = KC\), \(Q\) unchanged:
\(U' = \dfrac{Q^2}{2KC} = \dfrac{U}{K}\). Energy becomes \(\dfrac{1}{K}\) times.
(ii) Electric field \(E' = \dfrac{E}{K}\). Field becomes \(\dfrac{1}{K}\) times its initial value.
Question 52
(a) Obtain the expression for the energy stored per unit volume in a charged parallel plate capacitor. (b) The electric field inside a parallel plate capacitor is E. Find the amount of work done in moving a charge q over a closed rectangular loop abcda.
\(U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}\dfrac{\varepsilon_0 A}{d}(Ed)^2 = \dfrac{1}{2}\varepsilon_0 E^2 \times Ad\)
Energy density \(u = \dfrac{U}{Ad} =\) \(\dfrac{1}{2}\varepsilon_0 E^2\)
(b) The work done in moving a charge \(q\) over a closed rectangular loop \(abcda\) is zero, because the net displacement is zero.
\(W = q\vec{E}\cdot\vec{d} = 0\)
Question 53
Derive an expression for the energy stored in a parallel plate capacitor. On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of \(\varepsilon_r = 10\) is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the (i) capacitance, (ii) electric field, and (iii) energy density of the capacitor change.
Work done to supply charge \(dq\): \(dW = \dfrac{q}{C}\,dq\). Total work: \(W = \dfrac{Q^2}{2C}\).
Energy stored: \(U = \dfrac{1}{2}CV^2\)
New conditions: \(d' = d/2\), \(K = 10\), battery connected so \(V\) is constant.
(i) \(C' = \dfrac{K\varepsilon_0 A}{d/2} = \dfrac{20\varepsilon_0 A}{d} = 20C\). Capacitance becomes 20 times.
(ii) \(E' = \dfrac{V}{d/2} = \dfrac{2V}{d} = 2E\). Electric field doubles.
(iii) Energy density \(u' = \dfrac{1}{2}K\varepsilon_0 E'^2 = \dfrac{1}{2}(10)\varepsilon_0(2E)^2 = 40 \times \dfrac{1}{2}\varepsilon_0 E^2 = 40u\). Energy density becomes 40 times.
Question 54
A parallel plate capacitor, of capacitance 20 pF, is connected to a 100 V supply. After sometime the battery is disconnected, and the space between the plates of the capacitor is filled with a dielectric of dielectric constant 5. Calculate the energy stored in the capacitor (i) before (ii) after the dielectric has been put in between its plates.
(ii) After: battery disconnected so \(Q\) constant. \(C' = KC = 5 \times 20 = 100\ \text{pF}\).
\(U' = \dfrac{U}{K} = \dfrac{1 \times 10^{-7}}{5} =\) \(2 \times 10^{-8}\ \text{J}\)
Question 55
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and potential difference across each capacitor.
\(U = \dfrac{1}{2}CV^2 = \dfrac{1}{2} \times 12 \times 10^{-12} \times 2500 =\) \(1.5 \times 10^{-8}\ \text{J}\)
Series combination: \(C_s = \dfrac{12 \times 6}{12 + 6} = 4\ \text{pF}\)
Charge on each: \(q = C_sV = 4 \times 10^{-12} \times 50 =\) \(2 \times 10^{-10}\ \text{C}\)
P.d. across 12 pF: \(V_1 = \dfrac{q}{C_1} = \dfrac{2 \times 10^{-10}}{12 \times 10^{-12}} = 16.67\ \text{V}\)
P.d. across 6 pF: \(V_2 = \dfrac{q}{C_2} = \dfrac{2 \times 10^{-10}}{6 \times 10^{-12}} = 33.33\ \text{V}\)
Question 56
Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now? Also find the charge drawn from the battery in each case.
\(U_s = \dfrac{1}{2}C_sV^2 = \dfrac{1}{2} \times 6 \times 10^{-12} \times 2500 =\) \(7.5 \times 10^{-9}\ \text{J}\)
Charge \(Q_s = C_sV = 6 \times 10^{-12} \times 50 =\) \(3 \times 10^{-10}\ \text{C}\)
Parallel: \(C_p = 24\ \text{pF}\)
\(U_p = \dfrac{1}{2}C_pV^2 = \dfrac{1}{2} \times 24 \times 10^{-12} \times 2500 =\) \(3 \times 10^{-8}\ \text{J}\)
Charge \(Q_p = C_pV = 24 \times 10^{-12} \times 50 =\) \(1.2 \times 10^{-9}\ \text{C}\)
Question 57
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.
\(U_i = \dfrac{1}{2}CV^2 + \dfrac{1}{2}CV^2 = CV^2\)
After: Switch opened, charge on each capacitor is fixed at \(Q = CV\). Dielectric K inserted: \(C' = KC\).
Energy in each: \(U' = \dfrac{Q^2}{2C'} = \dfrac{(CV)^2}{2KC} = \dfrac{CV^2}{2K}\)
Total final energy: \(U_f = 2 \times \dfrac{CV^2}{2K} = \dfrac{CV^2}{K}\)
Ratio: \(\dfrac{U_i}{U_f} = \dfrac{CV^2}{CV^2/K} =\) \(K:1\)
Capacitor Networks and Calculations
Question 58
Three identical capacitors \(C_1\), \(C_2\) and \(C_3\) of capacitance 6 µF each are connected to a 12 V battery. Find (i) charge on each capacitor, (ii) equivalent capacitance of the network, (iii) energy stored in the network of capacitors.
(i) P.d. across \(C_3 = 12\ \text{V}\), so \(Q_3 = C_3V = 6 \times 10^{-6} \times 12 = 72\ \mu\text{C}\)
P.d. across \(C_4 = 12\ \text{V}\), so \(Q_4 = 3 \times 10^{-6} \times 12 = 36\ \mu\text{C}\)
Charge on \(C_1\) and \(C_2\) is also \(36\ \mu\text{C}\) each.
(ii) \(C_{eq} = C_4 + C_3 = 3 + 6 = 9\ \mu\text{F}\)
(iii) \(U = \dfrac{1}{2}C_{eq}V^2 = \dfrac{1}{2} \times 9 \times 10^{-6} \times 144 =\) \(6.48 \times 10^{-4}\ \text{J}\)
Question 59
The equivalent capacitance of the combination between A and B is 4 µF. (i) Calculate capacitance of the capacitor C. (ii) Calculate charge on each capacitor if a 12 V battery is connected across terminals A and B. (iii) What will be the potential drop across each capacitor?
\(\dfrac{1}{4} = \dfrac{1}{20} + \dfrac{1}{C} \Rightarrow \dfrac{1}{C} = \dfrac{1}{4} - \dfrac{1}{20} = \dfrac{4}{20} = \dfrac{1}{5} \Rightarrow C = 5\ \mu\text{F}\)
(ii) Total charge \(Q = C_{eq}V = 4 \times 10^{-6} \times 12 = 48\ \mu\text{C}\). Each capacitor carries \(48\ \mu\text{C}\).
(iii) P.d. across 20 µF: \(V_1 = \dfrac{48}{20} = 2.4\ \text{V}\)
P.d. across 5 µF: \(V_2 = \dfrac{48}{5} = 9.6\ \text{V}\)
Question 60
Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of \(\varepsilon_r = 4\). (i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 µF. (ii) Calculate the potential difference between the plates of X and Y. (iii) What is the ratio of electrostatic energy stored in X and Y?
(i) In series: \(\dfrac{1}{4} = \dfrac{1}{C} + \dfrac{1}{4C} = \dfrac{5}{4C} \Rightarrow C = 5\ \mu\text{F}\). So \(C_x = 5\ \mu\text{F}\), \(C_y = 20\ \mu\text{F}\).
(ii) Since capacitances are in ratio 1:4, potential drops are in ratio 4:1. Total 12 V, so \(V_x = \dfrac{4}{5} \times 12 = 9.6\ \text{V}\), \(V_y = \dfrac{1}{5} \times 12 = 2.4\ \text{V}\).
(iii) Since they carry the same charge, \(U = \dfrac{Q^2}{2C} \propto \dfrac{1}{C}\). Ratio \(U_x : U_y = C_y : C_x = 20 : 5 =\) \(4:1\)
Question 61
(i) Find equivalent capacitance between A and B in the combination given below. Each capacitor is of 2 µF capacitance. (ii) If a dc source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy stored in the network?
Total: \(C_{eq} = \dfrac{2}{3} + 1 = \dfrac{5}{3}\ \mu\text{F}\)
(ii) \(Q = C_{eq}V = \dfrac{5}{3} \times 10^{-6} \times 7 =\) \(1.17 \times 10^{-5}\ \text{C}\)
\(U = \dfrac{1}{2}C_{eq}V^2 = \dfrac{1}{2} \times \dfrac{5}{3} \times 10^{-6} \times 49 =\) \(4.08 \times 10^{-5}\ \text{J}\)
Question 62
Calculate the potential difference and the energy stored in the capacitor \(C_2\) in the circuit shown. Given potential at A is 90 V, \(C_1 = 20\ \mu\text{F}\), \(C_2 = 30\ \mu\text{F}\), \(C_3 = 15\ \mu\text{F}\).
\(\dfrac{1}{C} = \dfrac{1}{20} + \dfrac{1}{30} + \dfrac{1}{15} = \dfrac{3 + 2 + 4}{60} = \dfrac{9}{60} \Rightarrow C = \dfrac{60}{9} = \dfrac{20}{3}\ \mu\text{F}\)
Total charge \(Q = CV = \dfrac{20}{3} \times 10^{-6} \times 90 = 6 \times 10^{-4}\ \text{C}\)
P.d. across \(C_2\): \(V_2 = \dfrac{Q}{C_2} = \dfrac{6 \times 10^{-4}}{30 \times 10^{-6}} = 20\ \text{V}\)
Energy in \(C_2\): \(U_2 = \dfrac{1}{2}C_2V_2^2 = \dfrac{1}{2} \times 30 \times 10^{-6} \times 400 =\) \(6 \times 10^{-3}\ \text{J}\)
Question 63
Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of \(\varepsilon_r = 4\). (i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 pF. (ii) Calculate the potential difference between the plates of X and Y. (iii) Estimate the ratio of electrostatic energy stored in X and Y.
(i) Let \(C_x = C\), \(C_y = 4C\). \(\dfrac{1}{4} = \dfrac{1}{C} + \dfrac{1}{4C} = \dfrac{5}{4C} \Rightarrow C = 5\ \text{pF}\). So \(C_x = 5\ \text{pF}\), \(C_y = 20\ \text{pF}\).
(ii) For a given total voltage \(V\): \(V_x = \dfrac{4}{5}V\), \(V_y = \dfrac{1}{5}V\).
(iii) \(U_x : U_y = C_y : C_x = 20 : 5 =\) \(4:1\)
Question 64
A network of four capacitors each of 12 µF capacitance is connected to a 500 V supply. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.
(a) \(C_{eq} = 4 + 12 = 16\ \mu\text{F}\)
(b) Charge on \(C_4\): \(Q_4 = C_4V = 12 \times 10^{-6} \times 500 = 6000\ \mu\text{C}\)
Charge on each of the three series capacitors: \(Q_s = C_sV = 4 \times 10^{-6} \times 500 = 2000\ \mu\text{C}\)
Question 65
A network of four capacitors each of 15 µF capacitance is connected to a 500 V supply. Determine (a) equivalent capacitance of the network and (b) charge on each capacitor.
(a) \(C_{eq} = 5 + 15 = 20\ \mu\text{F}\)
(b) Charge on \(C_4\): \(Q_4 = 15 \times 10^{-6} \times 500 = 7500\ \mu\text{C}\)
Charge on each of the three series capacitors: \(Q_s = 5 \times 10^{-6} \times 500 = 2500\ \mu\text{C}\)
Capacitance Calculations
Question 66
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 360 µC. When potential across the capacitor is reduced by 120 V, the charge stored in it becomes 120 µC. Calculate: (i) the potential V and the unknown capacitance C, (ii) the charge stored in the capacitor if the voltage applied had increased by 120 V.
\(Q_2 = C(V - 120) \Rightarrow 120\ \mu\text{C} = C(V - 120)\) ...(ii)
Dividing (i) by (ii): \(3 = \dfrac{V}{V - 120} \Rightarrow 3V - 360 = V \Rightarrow 2V = 360 \Rightarrow V = 180\ \text{V}\)
From (ii): \(120 \times 10^{-6} = C(180 - 120) = 60C \Rightarrow C = 2\ \mu\text{F}\)
(ii) \(Q' = C(V + 120) = 2 \times 10^{-6} \times 300 =\) \(600\ \mu\text{C}\)
Question 67
A capacitor of unknown capacitance is connected across a battery of V volts. The charge stored in it is 300 µC. When potential across the capacitor is reduced by 100 V, the charge stored in it becomes 100 µC. Calculate the potential V and the unknown capacitance. What will be the charge stored in the capacitor if the voltage applied had increased by 100 V?
Dividing: \(3 = \dfrac{V}{V - 100} \Rightarrow 3V - 300 = V \Rightarrow 2V = 300 \Rightarrow V = 150\ \text{V}\)
From (ii): \(100 = C(150 - 100) = 50C \Rightarrow C = 2\ \mu\text{F}\)
\(Q' = C(V + 100) = 2 \times 10^{-6} \times 250 =\) \(500\ \mu\text{C}\)
Question 68
A capacitor of unknown capacitance is connected across a battery of V volts. A charge of 360 pC is stored in it. When the potential across the capacitor is reduced by 120 V, the charge stored becomes 120 pC. Calculate V and the unknown capacitance. What would have been the charge on the capacitor if the voltage were increased by 120 V?
\(Q' = C(V + 120) = 2 \times 10^{-12} \times 300 =\) \(600\ \text{pC}\)
Question 69
A capacitor of unknown capacitance is connected across a battery of V volts. A charge of 240 pC is stored in it. When the potential across the capacitor is reduced by 80 V, the charge stored becomes 80 pC. Calculate V and the unknown capacitance. What would have been the charge in the capacitor if the voltage were increased by 80 V?
Dividing: \(3 = \dfrac{V}{V - 80} \Rightarrow 3V - 240 = V \Rightarrow 2V = 240 \Rightarrow V = 120\ \text{V}\)
From (ii): \(80 = C(120 - 80) = 40C \Rightarrow C = 2\ \text{pF}\)
\(Q' = C(V + 80) = 2 \times 10^{-12} \times 200 =\) \(400\ \text{pC}\)
Question 70
A capacitor of unknown capacitance is connected across a battery of V volts. A charge of 120 µC is stored in it. When the potential across the capacitor is reduced by 40 V, the charge stored becomes 40 µC. Calculate V and the unknown capacitance. What would have been the charge in the capacitor if the voltage were increased by 40 V?
Dividing: \(3 = \dfrac{V}{V - 40} \Rightarrow 3V - 120 = V \Rightarrow 2V = 120 \Rightarrow V = 60\ \text{V}\)
From (ii): \(40 = C(60 - 40) = 20C \Rightarrow C = 2\ \mu\text{F}\)
\(Q' = C(V + 40) = 2 \times 10^{-6} \times 100 =\) \(200\ \mu\text{C}\)
Question 71
Two capacitors of unknown capacitances \(C_1\) and \(C_2\) are connected first in series and then in parallel across a battery of 100 V. If the energy stored in the two combinations is 0.045 J and 0.25 J respectively, determine the value of \(C_1\) and \(C_2\). Also calculate the charge on each capacitor in parallel combination.
Parallel energy: \(U_p = \dfrac{1}{2}C_pV^2 \Rightarrow 0.25 = \dfrac{1}{2}C_p(100)^2 \Rightarrow C_p = 50\ \mu\text{F}\)
\(C_1 + C_2 = 50\) and \(\dfrac{C_1C_2}{C_1 + C_2} = 9 \Rightarrow C_1C_2 = 450\)
Solving: \(C_1 = 40\ \mu\text{F}\), \(C_2 = 10\ \mu\text{F}\) (or vice versa).
Charge on each in parallel: \(Q_1 = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3}\ \text{C}\), \(Q_2 = 10 \times 10^{-6} \times 100 = 1 \times 10^{-3}\ \text{C}\).
Mixed Concepts
Question 72
Two parallel plate capacitors of capacitances \(C_1\) and \(C_2\) such that \(C_1 = 3C_2\) are connected across a battery of V volts. Initially the key is closed to fully charge the capacitors. The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination, before and after the introduction of the dielectric slab.
(i) Before: \(C_i = 4C\). After: each becomes \(KC\), so \(C_f = 4KC\). Ratio \(C_i : C_f = 1 : K\).
(ii) Energy before: \(U_i = \dfrac{1}{2}(4C)V^2 = 2CV^2\)
After: since key is open, total charge is constant. Initial total charge \(Q = 4CV\). Final capacitance \(4KC\), so \(U_f = \dfrac{(4CV)^2}{2(4KC)} = \dfrac{16C^2V^2}{8KC} = \dfrac{2CV^2}{K}\)
Ratio \(U_i : U_f =\) \(K : 1\)
Question 73
Two parallel plate capacitors of capacitances \(C_1\) and \(C_2\) such that \(C_1 = 2C_2\) are connected across a battery of V volts. Initially the key is closed to fully charge the capacitors. The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination, before and after the introduction of the dielectric slab.
(i) After: \(C_f = 3KC\). Ratio \(C_i : C_f = 1 : K\).
(ii) \(U_i = \dfrac{1}{2}(3C)V^2 = \dfrac{3}{2}CV^2\). \(U_f = \dfrac{(3CV)^2}{2(3KC)} = \dfrac{9C^2V^2}{6KC} = \dfrac{3CV^2}{2K}\).
Ratio \(U_i : U_f =\) \(K : 1\)
Question 74
Two parallel plate capacitors of capacitances \(C_1\) and \(C_2\) such that \(C_1 = C_2/2\) are connected across a battery of V volts. Initially the key is closed to fully charge the capacitors. The key is now thrown open and a dielectric slab of dielectric constant K is inserted in the two capacitors to completely fill the gap between the plates. Find the ratio of (i) the net capacitance and (ii) the energies stored in the combination, before and after the introduction of the dielectric slab.
(i) After: \(C_f = 3KC\). Ratio \(C_i : C_f = 1 : K\).
(ii) Same as Question 73: \(U_i : U_f = K : 1\).
Question 75
A parallel plate capacitor of capacitance C is charged to a potential V by a battery. Without disconnecting the battery, the distance between the plates is tripled and a dielectric medium of k = 10 is introduced between the plates. Explain giving reasons, how will the following be affected: (i) capacitance of the capacitor, (ii) charge on the capacitor, and (iii) energy density of the capacitor.
(i) \(C' = \dfrac{K\varepsilon_0 A}{3d} = \dfrac{10}{3} \cdot \dfrac{\varepsilon_0 A}{d} = \dfrac{10C}{3}\). Capacitance becomes \(\dfrac{10}{3}\) times.
(ii) \(Q' = C'V = \dfrac{10}{3}CV = \dfrac{10Q}{3}\). Charge becomes \(\dfrac{10}{3}\) times.
(iii) \(E' = \dfrac{V}{3d} = \dfrac{E}{3}\). Energy density \(u' = \dfrac{1}{2}K\varepsilon_0 E'^2 = \dfrac{1}{2}(10)\varepsilon_0\left(\dfrac{E}{3}\right)^2 = \dfrac{10}{9}u\). Energy density becomes \(\dfrac{10}{9}\) times.
Question 76
Two metallic spheres of radii R and 2R are charged so that both have the same surface charge density \(\sigma\). If they are connected to each other with a conducting wire, in which direction will the charge flow and why?
Potential of sphere of radius 2R: \(V_{2R} = \dfrac{kQ_2}{2R} = \dfrac{k(\sigma \cdot 4\pi (2R)^2)}{2R} = 8\pi k\sigma R\)
Since \(V_{2R} > V_R\), charge will flow from the bigger sphere to the smaller sphere.
Question 77
(a) Explain, using suitable diagrams, the difference in the behaviour of a (i) conductor and (ii) dielectric in the presence of an external electric field. Define the terms polarization of a dielectric and write its relation with susceptibility. (b) A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge Q/2 is placed at its centre C and another charge +2Q is placed outside the shell at a distance x from the centre. Find (i) the force on the charge at the centre of the shell and at the point A, (ii) the electric flux through the shell.
(ii) In a dielectric, the external field induces a net dipole moment by stretching/reorienting molecules. The induced field opposes but does not cancel the external field.
Polarization: Induced dipole moment per unit volume. For linear isotropic dielectrics, \(P = \chi_e \varepsilon_0 E\), where \(\chi_e\) is susceptibility.
(b) (i) Force on the charge at the centre is zero (shielded by the shell).
Force on the charge 2Q at point A: the shell behaves as if all its charge Q (plus the enclosed Q/2) is at the centre. Total effective charge \(= Q + Q/2 = 3Q/2\).
\(F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{(3Q/2)(2Q)}{x^2} = \dfrac{3Q^2}{4\pi\varepsilon_0 x^2}\)
(ii) Flux through the shell = enclosed charge/ε₀ = \(\dfrac{Q/2}{\varepsilon_0} = \dfrac{Q}{2\varepsilon_0}\)
Question 78
(i) If two similar large plates, each of area A having surface charge densities \(+\sigma\) and \(-\sigma\) are separated by a distance d in air, find the expressions for (a) field at points between the two plates and on outer side of the plates. Specify the direction of the field in each case. (b) the potential difference between the plates. (c) the capacitance of the capacitor so formed.
Outside the plates, the fields cancel: \(E = 0\).
(b) Potential difference: \(V = Ed = \dfrac{\sigma d}{\varepsilon_0}\)
(c) Capacitance: \(C = \dfrac{Q}{V} = \dfrac{\sigma A}{\sigma d/\varepsilon_0} = \dfrac{\varepsilon_0 A}{d}\)
Question 79
(a) Obtain the expression for the potential due to a point charge. (b) Use the above expression to show that the potential, due to an electric dipole (length 2a), varies as the inverse square of the distance r of the field point from the centre of the dipole for r > a.
\(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}\)
(b) For a dipole with charges \(+q\) and \(-q\) separated by \(2a\), the potential at a point P at distance r (making angle \(\theta\) with the dipole axis) is:
\(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r_1} + \dfrac{1}{4\pi\varepsilon_0}\dfrac{(-q)}{r_2}\)
For \(r \gg a\), using binomial approximation: \(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}\). This shows \(V \propto \dfrac{1}{r^2}\) — inverse square of distance.
Question 80
(a) Define the SI unit of capacitance. (b) Obtain the expression for the capacitance of a parallel plate capacitor. (c) Derive the expression for the effective capacitance of a series combination of n capacitors.
(b) \(C = \dfrac{\varepsilon_0 A}{d}\) (derivation same as Question 50).
(c) In series, charge on each capacitor is the same (Q). Total potential difference:
\(V = V_1 + V_2 + \ldots + V_n = \dfrac{Q}{C_1} + \dfrac{Q}{C_2} + \ldots + \dfrac{Q}{C_n}\)
\(\dfrac{V}{Q} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots + \dfrac{1}{C_n}\)
\(\dfrac{1}{C_s} = \sum_{i=1}^{n} \dfrac{1}{C_i}\)
Question 81
(a) Write two properties by which electric potential is related to the electric field. (b) Two point charges \(q_1\) and \(q_2\), separated by a distance of \(r_{12}\), are kept in an external electric field. Derive an expression for the potential energy of the system of two charges in the field.
(ii) The magnitude of electric field is given by the change in magnitude of potential per unit displacement normal to a charged conducting surface: \(E = -\dfrac{dV}{dr}\).
(b) Work done in bringing \(q_1\) to a point against external field: \(W_1 = q_1V(\vec{r}_1)\)
Work done in bringing \(q_2\) against the external field and the field due to \(q_1\): \(W_2 = q_2V(\vec{r}_2) + \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}\)
Total work = potential energy: \(U = q_1V(\vec{r}_1) + q_2V(\vec{r}_2) + \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}\)
Question 82
(a) Obtain the expression for the potential due to an electric dipole of dipole moment p at a point V on the axial line. (b) Two identical capacitors of plate dimensions l × b and plate separation d have dielectric slabs filled in between the space of the plates. Obtain the relation between the dielectric constants K, K₁ and K₂.
\(V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2 - a^2} \approx \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}\) for \(r \gg a\)
(b) If the two slabs fill the space (each half), the two capacitors are in parallel. One has dielectric K₁ and the other K₂, both with area \(l \times b/2\).
\(C_1 = \dfrac{K_1\varepsilon_0 (lb/2)}{d}\), \(C_2 = \dfrac{K_2\varepsilon_0 (lb/2)}{d}\)
Equivalent \(C = C_1 + C_2 = \dfrac{\varepsilon_0 lb}{2d}(K_1 + K_2)\)
If a single slab of constant K fills the whole space: \(C = \dfrac{K\varepsilon_0 lb}{d}\). Equating: \(K = \dfrac{K_1 + K_2}{2}\)
Equipotential Surfaces and Field Relation
Question 83
(a) Write two properties of equipotential surfaces. Depict equipotential surfaces due to an isolated point charge. Why do the equipotential surfaces get closer as the distance between the equipotential surface and the source charge decreases? (b) An electric dipole of dipole moment \(\vec{p}\) is placed in a uniform electric field \(\vec{E}\). Deduce the expression for the torque acting on it.
(i) No work is done in moving a test charge over an equipotential surface.
(ii) No two equipotential surfaces can intersect each other.
(iii) Equipotential surfaces due to an isolated point charge are spherical.
(iv) The electric field at every point is normal to the equipotential surface passing through that point.
Since \(dV = -E\,dr\), for the same \(dV\), \(dr \propto \dfrac{1}{E}\). As \(E\) is larger closer to the charge, \(dr\) is smaller — so surfaces get closer.
(b) Torque on dipole: \(\tau = pE\sin\theta\), vector form \(\vec{\tau} = \vec{p} \times \vec{E}\).
Question 84
(a) Write two properties by which electric potential is related to the electric field. (b) Two point charges \(q_1\) and \(q_2\), separated by a distance of \(r_{12}\), are kept in an external electric field. Derive an expression for the potential energy of the system of two charges in the field.
Key Formulas to Remember
| Concept | Formula |
|---|---|
| Electric potential due to point charge | \(V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{Q}{r}\) |
| Potential due to dipole (axial) | \(V = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{p}{r^2 - a^2}\) |
| Potential due to dipole (equatorial) | \(V = 0\) |
| Potential energy of dipole | \(U = -pE\cos\theta\) |
| Relation between E and V | \(E = -\dfrac{dV}{dr}\) |
| Capacitance of parallel plate | \(C = \dfrac{\varepsilon_0 A}{d}\) (air), \(C = \dfrac{K\varepsilon_0 A}{d}\) (dielectric) |
| Capacitors in series | \(\dfrac{1}{C_s} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \ldots\) |
| Capacitors in parallel | \(C_p = C_1 + C_2 + \ldots\) |
| Energy stored in capacitor | \(U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} = \dfrac{1}{2}QV\) |
| Energy density | \(u = \dfrac{1}{2}\varepsilon_0 E^2\) |
| Potential energy of two charges | \(U = \dfrac{1}{4\pi\varepsilon_0} \cdot \dfrac{q_1q_2}{r}\) |
FAQs
Are these questions enough for CBSE Class 12 Physics Chapter 2?
Yes. These questions cover every major topic — potential due to a dipole, equipotential surfaces, capacitance of parallel plate capacitors, combinations of capacitors, and energy stored — that CBSE has asked in previous board exams. Combine them with NCERT exercises and you'll be well prepared.
Which topics are most important for board exams?
Derivation of potential due to a dipole (axial and equatorial), energy stored in a capacitor, capacitors in series and parallel, effect of dielectric on capacitance, and equipotential surfaces are consistently high-weightage topics.
How should I prepare for numericals in this chapter?
Practise capacitance calculations with series-parallel networks, energy calculations with and without dielectrics, and problems involving unknown capacitance (using Q = CV with changed voltage). Focus on understanding when battery is connected vs disconnected — it decides whether V or Q stays constant.