CBSE Class 12 Physics Electrostatics Subjective Problems

CBSE Class 12 Physics Electrostatics Subjective Problems. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

CLASS 12 PHYSICS : ELECTROSTATICS
SUBJECTIVE PROBLEMS

1. Two equal point charges +q are separated by a distance 2a. A point test charge is located in a plane which is normal to the line joining these charges and midway between them. Find radius r of the circle of symmetry in this plane for which the force on the test charge has a maximum value.

Answer: r = \( \frac{a}{\sqrt{2}} \)
1. A point on the circle is at distance √(r² + a²) from each charge. The components of the two fields along the line joining the charges cancel, and the components along the radius add.
2. E = 2 \( \frac{kq}{r^{2} + a^{2}} \) × \( \frac{r}{\sqrt{r^{2} + a^{2}}} \) = \( \frac{2kq r}{(r^{2} + a^{2})^{3/2}} \).
3. For maximum E, \( \frac{dE}{dr} \) = 0: (r² + a²)^3/2 - r × 3r (r² + a²)^1/2 = 0.
4. r² + a² = 3r², so 2r² = a².
5. r = \( \frac{a}{\sqrt{2}} \).

Teacher's Note:
a) The force on the test charge is q₀E, so maximising the force means maximising the field.
b) Check that E falls to zero at r = 0 and at very large r, so the stationary point must be a maximum.

2. Three charge q, 2q and 8q are to be placed on 9 cm long straight line. Find the position where the charges should be placed such that the potential energy of this system is minimum. Find the electric field at the position of charge q due to the other two charges.

Answer: 2q and 8q at the two ends, q at 3 cm from 2q; the electric field at q is zero.
1. All three charges are of the same sign and repel one another, so the two outer charges should be at the two ends of the 9 cm line, and the third charge lies between them.
2. Energy in units of kq² per cm: with 2q in the middle the minimum is about 4.15; with 8q in the middle it is about 5.40; with q in the middle, 2q and 8q at the ends, the energy is lowest.
3. Let q be at x cm from 2q. U = kq²[\( \frac{16}{9} \) + \( \frac{2}{x} \) + \( \frac{8}{9 - x} \)].
4. \( \frac{dU}{dx} \) = 0 gives \( \frac{2}{x^{2}} \) = \( \frac{8}{(9 - x)^{2}} \), so 9 - x = 2x and x = 3 cm.
5. So 2q and 8q are at the ends 9 cm apart, and q is 3 cm from 2q (6 cm from 8q).
6. Electric field at q: due to 2q = \( \frac{k \cdot 2q}{(3)^{2}} \) = \( \frac{2kq}{9} \) (away from 2q); due to 8q = \( \frac{k \cdot 8q}{(6)^{2}} \) = \( \frac{2kq}{9} \) (away from 8q). These are in opposite directions, so the net field is zero.

Teacher's Note:
a) At the position of minimum potential energy the charge q is in equilibrium, which is why the field there is zero.
b) Charges of the same sign cannot be brought closer without increasing the energy, so the largest charges are kept far apart.

3. A charge 2.0 μC is placed at the corner A of a square ABCD of side 0.1 m. A charge -1.0 μC is located at the centre of the square. Find the work done in carrying a charge of 0.3 μC from the corner C to B of the square.

Answer: W ≈ 1.58 × 10⁻² J
1. AC = 0.1√2 m = 0.1414 m, and the distance of the centre from each corner = 0.0707 m. AB = 0.1 m.
2. V_C = 9 × 10⁹ [\( \frac{2 \times 10^{-6}}{0.1414} \) - \( \frac{1 \times 10^{-6}}{0.0707} \)] = 9 × 10⁹ [1.414 × 10⁻⁵ - 1.414 × 10⁻⁵] = 0.
3. V_B = 9 × 10⁹ [\( \frac{2 \times 10^{-6}}{0.1} \) - \( \frac{1 \times 10^{-6}}{0.0707} \)] = 9 × 10⁹ × 5.86 × 10⁻⁶ ≈ 5.27 × 10⁴ V.
4. Work done = q₀ (V_B - V_C) = 0.3 × 10⁻⁶ × 5.27 × 10⁴.
5. Work done ≈ 1.58 × 10⁻² J.

Teacher's Note:
a) Work done in moving a charge depends only on the potential difference between the end points, not on the path.
b) Corners B and D are at the same potential as each other by symmetry, but not as C.

4. An infinite number of charges, each equal to q Coulomb, are placed along the x-axis at x = 1, 2, 4, 8......... and so on where x is in metres. Find the potential and electric field at the point x = 0 due to this set of charges. Find the potential and electric field in the above set if the consecutive charges have opposite sign.

Answer: Same sign: V = \( \frac{q}{2π ε_0} \), E = \( \frac{q}{3π ε_0} \). Alternate sign: V = \( \frac{q}{6π ε_0} \), E = \( \frac{q}{5π ε_0} \).
1. All charges of the same sign: V = kq(1 + \( \frac{1}{2} \) + \( \frac{1}{4} \) + \( \frac{1}{8} \) + ...) = kq × 2 = \( \frac{q}{2π ε_0} \).
2. E = kq(1 + \( \frac{1}{4} \) + \( \frac{1}{16} \) + ...) = kq × \( \frac{4}{3} \) = \( \frac{q}{3π ε_0} \), directed along the negative x-axis (away from the positive charges).
3. Alternate signs: V = kq(1 - \( \frac{1}{2} \) + \( \frac{1}{4} \) - ...) = kq × \( \frac{1}{1 + 1/2} \) = kq × \( \frac{2}{3} \) = \( \frac{q}{6π ε_0} \).
4. E = kq(1 - \( \frac{1}{4} \) + \( \frac{1}{16} \) - ...) = kq × \( \frac{1}{1 + 1/4} \) = kq × \( \frac{4}{5} \) = \( \frac{q}{5π ε_0} \), again along the negative x-axis (the nearest charge, at x = 1, dominates).

Teacher's Note:
a) Both series are geometric: the sum of a + ar + ar² + ... is a/(1 - r), with r = ±1/2 for the potential and ±1/4 for the field.
b) The field is a vector and the potential a scalar, so the fields of charges at different distances are added with their directions.

5. Four charges +q, +q, -q and -q are placed respectively at the corners A, B, C and D of a square with side a, arranged in the given order. Calculate the electrical potential and intensity at O, the center of the square. If E and F are the mid-points of sides BC and CD, respectively, what will be the work done in carrying a charge e from O to E and from O to F?

Answer: V = 0 and E = \( \frac{\sqrt{2} q}{π ε_0 a^{2}} \) along OF at O; work from O to E = 0; work from O to F has magnitude \( \frac{(\sqrt{5} - 1) q e}{\sqrt{5} π ε_0 a} \).
1. The distance of each corner from O = \( \frac{a}{\sqrt{2}} \). Potential at O = \( \frac{kq}{a/\sqrt{2}} \)(+1 + 1 - 1 - 1) = 0.
2. Field at O: the fields of A and C are in the same direction along AC (each \( \frac{kq}{a^{2}/2} \) = \( \frac{2kq}{a^{2}} \)), and the fields of B and D are in the same direction along BD.
3. Resultant of the four fields = \( \frac{4\sqrt{2} kq}{a^{2}} \) = \( \frac{\sqrt{2} q}{π ε_0 a^{2}} \), directed from the midpoint of AB towards the midpoint of CD, that is along OF.
4. Potential at E (mid-point of BC): the distances from A, B, C, D are \( \frac{\sqrt{5} a}{2} \), \( \frac{a}{2} \), \( \frac{a}{2} \), \( \frac{\sqrt{5} a}{2} \), so V_E = kq[\( \frac{2}{\sqrt{5} a} \) + \( \frac{2}{a} \) - \( \frac{2}{a} \) - \( \frac{2}{\sqrt{5} a} \)] = 0.
5. Work done from O to E = e (V_E - V_O) = 0.
6. Potential at F (mid-point of CD): V_F = kq[\( \frac{4}{\sqrt{5} a} \) - \( \frac{4}{a} \)] = -\( \frac{4kq}{a} \)(1 - \( \frac{1}{\sqrt{5}} \)).
7. Work done by the external agent from O to F = e (V_F - V_O) = -\( \frac{4kqe}{a} \)(1 - \( \frac{1}{\sqrt{5}} \)) = -\( \frac{(\sqrt{5} - 1) q e}{\sqrt{5} π ε_0 a} \). The electric field does the same amount of work with the opposite sign, + \( \frac{(\sqrt{5} - 1) q e}{\sqrt{5} π ε_0 a} \).

Teacher's Note:
a) The potential is zero at O and E because positive and negative charges are at equal distances from them.
b) The printed key gives 4qe(√5 - 1)/a for the work from O to F; the exact potential at F carries a further factor 1/√5, as shown above.
c) The work done by the field equals the charge times the fall in potential.

6. A charge of 4 × 10⁻⁹ coulmb is uniformly distributed over the surface of a ring shaped conductor of radius 0.3 m. Calculate the intensity and direction of the electric field at a point on the axis of the ring at a distance of 0.4 m from the plane of the conductor. Find the intensity at the centre of ring.

Answer: E = 115.2 V/m along the axis, away from the ring; E = 0 at the centre.
1. For a point on the axis, E = \( \frac{k Q x}{(R^{2} + x^{2})^{3/2}} \).
2. R² + x² = 0.09 + 0.16 = 0.25, so (R² + x²)^3/2 = 0.125.
3. E = \( \frac{9 \times 10^{9} \times 4 \times 10^{-9} \times 0.4}{0.125} \) = \( \frac{14.4}{0.125} \) = 115.2 V/m.
4. The ring is positively charged, so the field is directed along the axis, away from the ring.
5. At the centre of the ring, the contributions of diametrically opposite elements cancel, so E = 0.

Teacher's Note:
a) By symmetry the components of the field perpendicular to the axis cancel, leaving only the axial component.
b) The field is zero at the centre and also tends to zero far away, with a maximum at x = R/√2.

7. Two circular loops of radii 0.05 m and 0.09 m respectively are placed vertically such that their axes coincide and their centres are 0.12 m apart. A charge of one micro-coulomb is spread uniformly on each loop. Find the potential difference between the centres of loops.

[Figure: Two circular loops placed vertically with a common horizontal axis. The smaller loop of radius 0.05 m has its centre at O and the larger loop of radius 0.09 m has its centre at P; the centres are 0.12 m apart. Lines are drawn from O to the rim of the larger loop and from P to the rim of the smaller loop]

Answer: Potential difference ≈ 71 kV (7.1 × 10⁴ V)
1. Let O be the centre of the 0.05 m loop and P the centre of the 0.09 m loop. Distance from O to the rim of the 0.09 m loop = √(0.12² + 0.09²) = 0.15 m. Distance from P to the rim of the 0.05 m loop = √(0.12² + 0.05²) = 0.13 m.
2. V_O = kQ(\( \frac{1}{0.05} \) + \( \frac{1}{0.15} \)) = 9 × 10⁹ × 10⁻⁶ × (20 + 6.67) = 2.4 × 10⁵ V.
3. V_P = kQ(\( \frac{1}{0.09} \) + \( \frac{1}{0.13} \)) = 9 × 10³ × (11.11 + 7.69) = 1.69 × 10⁵ V.
4. Potential difference V_O - V_P = 9 × 10³ × (26.67 - 18.80) ≈ 7.1 × 10⁴ V.

Teacher's Note:
a) Every element of a ring is at the same distance from a point on its axis, so the potential is simply kQ/distance.
b) Add the potentials due to both rings at each centre before subtracting.

8. A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards. With what period will the pendulum oscillate, if the electrostatic force acting on the sphere is less than the gravitational force ? (Assume that the oscillations are small).

Answer: T = 2π \( \sqrt{\frac{m l}{mg - qE}} \)
1. The electric force qE acts upward and the weight mg acts downward, with qE less than mg.
2. Net downward force = mg - qE, so the effective acceleration g_eff = \( \frac{mg - qE}{m} \).
3. Period T = 2π \( \frac{\sqrt{l}}{\sqrt{g_{eff}}} \).
4. T = 2π \( \sqrt{\frac{m l}{mg - qE}} \).

Teacher's Note:
a) The pendulum formula still holds, with the ordinary g replaced by the effective acceleration g - qE/m.
b) If qE were greater than mg the sphere would be pushed upward and the pendulum would not oscillate in the usual way.

9. A particle of mass 40 mg and carrying a charge of 5 × 10⁻⁹ C is moving directly towards a fixed positive point charge of magnitude 10⁻⁸ C. When it is at a distance of 10 cm from the fixed point charge, it has a velocity of 50 cm/second. Find the distance from the fixed point charge where the particle will momentarily come to rest.

Answer: r = 4.7 × 10⁻² m (about 4.7 cm)
1. Mass m = 40 mg = 4 × 10⁻⁵ kg and initial speed v = 0.5 m/s, so the kinetic energy = \( \frac{1}{2} \) × 4 × 10⁻⁵ × 0.25 = 5 × 10⁻⁶ J.
2. kq₁q₂ = 9 × 10⁹ × 5 × 10⁻⁹ × 10⁻⁸ = 4.5 × 10⁻⁷ J m.
3. Initial potential energy = \( \frac{4.5 \times 10^{-7}}{0.1} \) = 4.5 × 10⁻⁶ J.
4. At the stopping point the kinetic energy is zero: \( \frac{4.5 \times 10^{-7}}{r} \) = 5 × 10⁻⁶ + 4.5 × 10⁻⁶ = 9.5 × 10⁻⁶ J.
5. r = \( \frac{4.5 \times 10^{-7}}{9.5 \times 10^{-6}} \) = 4.7 × 10⁻² m.

Teacher's Note:
a) Use conservation of energy: the kinetic energy lost equals the gain in electrostatic potential energy.
b) Convert mg to kg and cm to m before substituting.

10. Three point charges 1C, 2C and 3C are placed at the corners of an equilateral triangle of side 1 m. If these charges were to be shifted to the corners of a smaller equilateral triangle of side 0.5 m as shown in figure, find the work done to move these charges.

[Figure: A large equilateral triangle of side 1 m with charges 1 C at the top corner, 2 C at the bottom-left corner and 3 C at the bottom-right corner, and a smaller equilateral triangle of side 0.5 m inside it. Arrows show the charges being moved from the large triangle corners to the small triangle corners]

Answer: W = 9.9 × 10¹⁰ J
1. Potential energy of the system U = k \( \frac{q_1 q_2 + q_2 q_3 + q_3 q_1}{s} \), where s is the side of the triangle.
2. q₁q₂ + q₂q₃ + q₃q₁ = 2 + 6 + 3 = 11 C².
3. Initial energy U_i = 9 × 10⁹ × \( \frac{11}{1} \) = 9.9 × 10¹⁰ J.
4. Final energy U_f = 9 × 10⁹ × \( \frac{11}{0.5} \) = 1.98 × 10¹¹ J.
5. Work done = U_f - U_i = 9.9 × 10¹⁰ J.

Teacher's Note:
a) The work done by the external agent equals the increase in the potential energy of the system.
b) Because the charges repel, positive work has to be done to bring them closer.

11. A pendulum bob of mass 80 mg and carrying a charge of 2 × 10⁻⁸ coulombs, is at rest in a horizontal, uniform electric field of 20,000 volt/m. Find the tension in the pendulum and the angle it makes with vertical.

Answer: T ≈ 8.8 × 10⁻⁴ N (about 0.0009 N) and θ ≈ 27°2′
1. Weight mg = 80 × 10⁻⁶ × 9.8 = 7.84 × 10⁻⁴ N (vertically downward).
2. Electric force qE = 2 × 10⁻⁸ × 20000 = 4 × 10⁻⁴ N (horizontal).
3. For equilibrium, T cos θ = mg and T sin θ = qE.
4. T = √((7.84 × 10⁻⁴)² + (4 × 10⁻⁴)²) ≈ 8.8 × 10⁻⁴ N.
5. tan θ = \( \frac{4}{7.84} \) = 0.51, so θ ≈ 27°2′.

Teacher's Note:
a) Resolve the tension into vertical and horizontal components that balance the weight and the electric force respectively.
b) Convert 80 mg to 80 × 10⁻⁶ kg.

12. A particle having mass 9 × 10⁻³¹ kg and negative charge 1.6 × 10⁻¹⁹ coulomb is projected horizontally with a velocity of 10⁵ m/sec into a region between two infinite, horizontal parallel plates of metal. The distance between the plates is 0.3 cm and the particle enters 0.1 cm below the top plate. The top and the bottom plates are connected, respectively, to the positive and negative terminals of a 30-volts battery. Find the components of the velocity of the particle just before it hits one of the plates.

Answer: v_x = 10⁵ m/s and v_y ≈ 1.9 × 10⁶ m/s (towards the top plate)
1. Electric field between the plates E = \( \frac{30}{0.003} \) = 10⁴ V/m, directed downward (from the positive top plate).
2. The particle is negatively charged, so the force is upward, towards the top plate. Acceleration a = \( \frac{eE}{m} \) = \( \frac{1.6 \times 10^{-19} \times 10^{4}}{9 \times 10^{-31}} \) = 1.78 × 10¹⁵ m/s².
3. The particle enters 0.1 cm = 10⁻³ m below the top plate, so it hits the top plate after moving 10⁻³ m vertically.
4. v_y² = 2 a s = 2 × 1.78 × 10¹⁵ × 10⁻³ = 3.56 × 10¹² so v_y ≈ 1.9 × 10⁶ m/s (upward).
5. The horizontal component stays constant: v_x = 10⁵ m/s.

Teacher's Note:
a) The motion is like a projectile motion: constant velocity along the plates and uniform acceleration across them.
b) Gravity is neglected because it is negligible compared with the electric force on an electron.

13. Two parallel plate capacitors of capacity C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the space between the plates of the capacitor C is completely filled with a material of dielectric constant K. Find the final potential difference across the capacitors.

Answer: V′ = \( \frac{3V}{K + 2} \)
1. Total charge before disconnecting = (C + 2C) V = 3CV, and this charge stays constant.
2. After the dielectric is inserted, the capacitance of the first capacitor becomes KC, so the total capacitance of the parallel combination = KC + 2C.
3. Final potential difference V′ = \( \frac{3CV}{KC + 2C} \) = \( \frac{3V}{K + 2} \).

Teacher's Note:
a) With the battery disconnected, the charge is conserved while the potential changes.
b) Check: for K = 1 the potential stays at V.

14. Two dielectric slabs of dielectric constant K₁ and K₂ have been filled equally in between the plates of two similar parallel plate capacitors of area A and separation d as shown in figure. Find the capacitance in each case (a) and (b). Also find the ratio of capacitance of (a) and (b) if the values of K₁ and K₂ are 2 and 3 respectively.

[Figure: (a) A parallel plate capacitor of plate area A and separation d in which the left half of the space is filled with a slab of dielectric constant K₁ and the right half with a slab of dielectric constant K₂, each slab spanning the whole separation d. (b) A similar capacitor in which a slab of dielectric constant K₁ lies above a slab of dielectric constant K₂, each of area A and half of the separation d]

Answer: C_a = \( \frac{ε_0 A (K_1 + K_2)}{2d} \); C_b = \( \frac{2ε_0 A}{d} \) \( \frac{K_1 K_2}{K_1 + K_2} \); ratio = 25 : 24
1. Case (a): the slabs lie side by side, each with area \( \frac{A}{2} \) and the full separation d, so the capacitors are in parallel.
2. C_a = \( \frac{K_1 ε_0 A/2}{d} \) + \( \frac{K_2 ε_0 A/2}{d} \) = \( \frac{ε_0 A (K_1 + K_2)}{2d} \).
3. Case (b): the slabs lie one above the other, each of thickness \( \frac{d}{2} \) and area A, so the capacitors are in series with C₁ = \( \frac{2K_1 ε_0 A}{d} \) and C₂ = \( \frac{2K_2 ε_0 A}{d} \).
4. C_b = \( \frac{C_1 C_2}{C_1 + C_2} \) = \( \frac{2ε_0 A}{d} \) \( \frac{K_1 K_2}{K_1 + K_2} \).
5. For K₁ = 2 and K₂ = 3: C_a = \( \frac{5 ε_0 A}{2d} \) and C_b = \( \frac{2 ε_0 A}{d} \) × \( \frac{6}{5} \) = \( \frac{12 ε_0 A}{5d} \).
6. \( \frac{C_a}{C_b} \) = \( \frac{5}{2} \) × \( \frac{5}{12} \) = \( \frac{25}{24} \), so the ratio is 25 : 24.

Teacher's Note:
a) Slabs spanning the full gap side by side are in parallel; slabs stacked across the gap are in series.
b) The ratio does not depend on A or d.

15. Two different capacitors C₁ and C₂ are connected in parallel such that the energy of the combination is 0.25 J when the difference of potentials between their terminals is 10 volts. When these condensers are connected in series, the energy is 0.05 J, when the same potential difference is applied. Find C₁ and C₂

Answer: C₁ = 3.62 × 10⁻³ F and C₂ = 1.38 × 10⁻³ F
1. In parallel: \( \frac{1}{2} \) (C₁ + C₂) (10)² = 0.25, so C₁ + C₂ = 5 × 10⁻³ F.
2. In series: \( \frac{1}{2} \) \( \frac{C_1 C_2}{C_1 + C_2} \) (10)² = 0.05, so \( \frac{C_1 C_2}{C_1 + C_2} \) = 10⁻³ F, giving C₁C₂ = 5 × 10⁻⁶ F².
3. C₁ and C₂ are the roots of C² - (5 × 10⁻³) C + 5 × 10⁻⁶ = 0.
4. C = \( \frac{5 \times 10^{-3} \pm \sqrt{25 \times 10^{-6} - 20 \times 10^{-6}}}{2} \) = \( \frac{5 \times 10^{-3} \pm 2.236 \times 10^{-3}}{2} \).
5. C₁ = 3.62 × 10⁻³ F and C₂ = 1.38 × 10⁻³ F.

Teacher's Note:
a) Use the sum and the product of the two capacitances to form a quadratic equation.
b) The printed key (3.61 × 10⁻³ F and 1.39 × 10⁻³ F) differs only in the third significant figure.

16. Find the potential difference between points A and B of the circuit as shown in the figure. The e.m.f. of the battery is 10 volts and each capacitor C is of 8μF

[Figure: A 10 V battery on the left. The top wire has a capacitor C, then a junction, then another capacitor C, then terminal A. From the junction, a vertical capacitor C goes down to the bottom wire. Another vertical capacitor C goes down from the wire just before terminal A to the bottom wire, which ends at terminal B]

Answer: V_A - V_B = 2 V
1. No current flows to the terminals A and B (open circuit), so the top-right capacitor and the right-hand vertical capacitor form a series pair of C and C connected across the middle vertical capacitor.
2. The series pair has capacitance \( \frac{C}{2} \), and this is in parallel with the middle capacitor C: \( \frac{3C}{2} \).
3. This is in series with the top-left capacitor C: \( \frac{C \times 3C/2}{C + 3C/2} \) = \( \frac{3C}{5} \), so the charge drawn from the battery = \( \frac{3C}{5} \) × 10 = 6C.
4. Voltage across the parallel part = \( \frac{6C}{3C/2} \) = 4 V.
5. The 4 V divides equally between the two equal capacitors in series, so the voltage across the right-hand vertical capacitor is 2 V.
6. Terminals A and B are connected across this capacitor, so V_A - V_B = 2 V.

Teacher's Note:
a) Capacitors with equal capacitance in series share the voltage equally.
b) The value of C (8 μF) cancels out in the end.

17. A parallel plate capacitor consists of two metal plates of area A and separation d. A slab of thickness t and dielectric constant K is inserted between plates with its faces parallel to the plates and having the same surface area as that of the plates. Find the capitance of the system. If K = 2, find the value of t/d for which the capacitance of the system will be 3/2 times that of the air capacitor. Find the ratio of the energy of the air capacitor to that of the capacitor with the dielectric inserted inside.

[Figure: A parallel plate capacitor with plate separation d. A dielectric slab K of thickness t is attached to the left plate, and the remaining gap d - t is filled with air]

Answer: C = \( \frac{ε_0 A}{d - t + t/K} \); t/d = \( \frac{2}{3} \); the energy ratio is 3 : 2 for constant charge.
1. The gap contains air of thickness (d - t) and the slab of thickness t, in series. C = \( \frac{ε_0 A}{(d - t) + t/K} \).
2. For K = 2: C = \( \frac{ε_0 A}{d - t/2} \).
3. The air capacitor has C₀ = \( \frac{ε_0 A}{d} \). For C = \( \frac{3}{2} \) C₀: \( \frac{d}{d - t/2} \) = \( \frac{3}{2} \), so 2d = 3d - \( \frac{3t}{2} \).
4. \( \frac{3t}{2} \) = d, so \( \frac{t}{d} \) = \( \frac{2}{3} \).
5. If the capacitor carries the same charge in both cases (it is isolated), the energy is U = \( \frac{Q^{2}}{2C} \), so \( \frac{U_{air}}{U_{dielectric}} \) = \( \frac{C}{C_0} \) = \( \frac{3}{2} \).

Teacher's Note:
a) The slab is equivalent to removing a thickness t of air and putting in a thickness t/K.
b) The energy ratio depends on whether the charge or the voltage is kept constant; the answer 3 : 2 is for constant charge.

18. Two parallel plate condensers A and B having capacities of 1μF and 5μF are charged separately to the same potential of 100 V. Now the positive plate of A is connected to the negative plate of B and negative plate of A is connected to the positive plate of B. Find the final charge on each condenser and total loss of electrical energy in the condensers.

Answer: Final charges: 66.7 μC on A and 333.3 μC on B; energy lost = 1.67 × 10⁻² J
1. Initial charges: Q_A = 1 × 100 = 100 μC and Q_B = 5 × 100 = 500 μC.
2. Connected with opposite polarity, the net charge on the combination = 500 - 100 = 400 μC, and the capacitors are in parallel with total capacitance 6 μF.
3. Common potential V′ = \( \frac{400}{6} \) = 66.7 V.
4. Final charge on A = 1 × 66.7 = 66.7 μC and on B = 5 × 66.7 = 333.3 μC (A has its polarity reversed). Charge of 400 μC has moved from A to B through the wires.
5. Initial energy = \( \frac{1}{2} \) × 6 × 10⁻⁶ × (100)² = 0.03 J.
6. Final energy = \( \frac{1}{2} \) × 6 × 10⁻⁶ × (66.7)² = 0.0133 J.
7. Loss of energy = 0.03 - 0.0133 = 1.67 × 10⁻² J.

Teacher's Note:
a) Opposite plates joined together means the charges partly neutralise, so only the difference of the charges remains.
b) The printed key lists ±4 × 10⁻⁴ C, which is the charge that flows through the connecting wires; the charges left on the capacitors are 66.7 μC and 333.3 μC.

19. Two batteries, each of 60 volt, are connected to condensers C₁ and C₂ of capacitance 2.0 μF and 3.0μF respectively through a switch as shown in the circuit diagram. Find the charge which will flow after shorting of switch through sections 1, 2, and 3 in the directions indicated in the figure.

[Figure: Two 60 V batteries in series on the left, with a switch S joining the junction between the batteries to the junction between two capacitors on the right: C₁ (top right) and C₂ (bottom right). Sections 1, 2 and 3 are the top wire (battery to C₁), the middle wire containing the switch, and the bottom wire (battery to C₂); arrows on all three point to the right, towards the capacitors]

Answer: Section 1: -24 μC; section 2: +60 μC; section 3: -36 μC
1. Before the switch is closed, C₁ and C₂ are in series across 120 V: C = \( \frac{2 \times 3}{2 + 3} \) = 1.2 μF and the charge on each = 1.2 × 120 = 144 μC.
2. After the switch is closed, each capacitor is directly across one 60 V battery: Q₁ = 2 × 60 = 120 μC and Q₂ = 3 × 60 = 180 μC.
3. Section 1 (top plate of C₁): charge flowing = 120 - 144 = -24 μC.
4. Section 3 (bottom plate of C₂): the charge on the plate changes from -144 μC to -180 μC, so the charge flowing = -36 μC.
5. Section 2 (the node joining C₁ and C₂): the charge changes from 0 to (-120 + 180) = +60 μC, so the charge flowing = 60 μC.

Teacher's Note:
a) The middle node is isolated, so the charge flowing through section 2 equals the net charge that appears on it.
b) A negative value means that the charge actually flows opposite to the arrow shown in the figure.

20. From the given figure, find the value of the capacitance C if the equivalent capacitance between points A and B is to be 1 μF. All the capacitances are in μF.

[Figure: A network between terminals A and B. From A, a capacitor C leads to a junction, from which a 1 μF capacitor continues along the top wire to the right. From the junction, a vertical 8 μF capacitor goes down to a second junction, to which two 2 μF capacitors are connected, each going to the bottom wire. At the right end of the top wire, a 4 μF capacitor goes down to the bottom wire, and a 6 μF capacitor in series with a 12 μF capacitor also goes from the top wire to the bottom wire. The bottom wire ends at B]

Answer: C = 2.3 μF (answer as given in the PDF key; the steps below follow one reading of the figure and give about 1.39 μF)
1. The two 2 μF capacitors on the lower left are both connected between the lower end of the 8 μF capacitor and the bottom wire, so they are in parallel: 4 μF.
2. This 4 μF is in series with the 8 μF: \( \frac{8 \times 4}{8 + 4} \) = \( \frac{8}{3} \) μF.
3. On the right, 6 μF and 12 μF are in series: \( \frac{6 \times 12}{18} \) = 4 μF; this is in parallel with the 4 μF capacitor: 8 μF.
4. This 8 μF is in series with the 1 μF capacitor: \( \frac{1 \times 8}{9} \) = \( \frac{8}{9} \) μF.
5. The two paths from the junction after C to the bottom wire are in parallel: \( \frac{8}{3} \) + \( \frac{8}{9} \) = \( \frac{32}{9} \) μF.
6. This is in series with C: \( \frac{1}{C} \) + \( \frac{9}{32} \) = 1, so \( \frac{1}{C} \) = \( \frac{23}{32} \).
7. C = \( \frac{32}{23} \) ≈ 1.39 μF.

Teacher's Note:
a) Redraw the network and mark the two junctions that every capacitor connects to before combining.
b) The answer given in the key is 2.3 μF.

21. Two insulated metallic solid spheres of radii R and 2R are charged such that both of them have same charge density σ. The spheres are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

Answer: New charge density on the bigger sphere = \( \frac{5σ}{6} \)
1. Charges: Q₁ = σ 4πR² and Q₂ = σ 4π(2R)² = 4σ 4πR², so the total charge = 5σ 4πR².
2. After connection both spheres have the same potential: \( \frac{kQ_1'}{R} \) = \( \frac{kQ_2'}{2R} \), so Q₂′ = 2Q₁′.
3. Q₁′ + Q₂′ = 5σ 4πR², so Q₂′ = \( \frac{2}{3} \) × 5σ 4πR².
4. New density on the bigger sphere = \( \frac{Q_2'}{4π (2R)^{2}} \) = \( \frac{2}{3} \) × \( \frac{5σ 4πR^{2}}{4π \cdot 4R^{2}} \) = \( \frac{5σ}{6} \).

Teacher's Note:
a) Connected conductors reach the same potential, not the same charge.
b) The charge on a sphere at a fixed potential is proportional to its radius.

22. Two capacitors A and B with capacities 3μF and 2μF are charged to potential differences of 100 V and 180 V respectively. The plates of the capacitors are then connected as shown in the figure with one wire from each capacitor free. The upper plate of A is positive and that of B is negative. An uncharged 2μF capacitor C with lead wires falls on the free ends to complete the circuit. Calculate
(a) the final charge on the three capacitors,
(b) the amount of electrostatic energy stored in the system before and after the completion of the circuit.

[Figure: Capacitor A (3 μF, 100 V) and capacitor B (2 μF, 180 V) side by side. The upper plate of A is positive and that of B is negative (the lower plate of B is positive). Their lower plates are joined by a wire along the bottom. An uncharged 2 μF capacitor C is connected across the top between the upper plates of A and B]

Answer: (a) q_A = 90 μC, q_B = 150 μC, q_C = 210 μC; (b) energy before = 4.74 × 10⁻² J and after = 1.8 × 10⁻² J
1. Initial charges: Q_A = 3 × 100 = 300 μC (upper plate +) and Q_B = 2 × 180 = 360 μC (upper plate -).
2. After connecting C, let the charge on the left plate of C be x. The isolated section joining the upper plate of A and the left plate of C keeps its charge: q_A + x = 300.
3. The section joining the right plate of C and the upper plate of B: -x + q_B = -360, so q_B = x - 360 (as the signed charge on its upper plate).
4. The sum of the potential drops round the loop is zero: \( \frac{q_A}{3} \) - \( \frac{x}{2} \) = \( \frac{q_B}{2} \).
5. Substituting: \( \frac{300 - x}{3} \) - \( \frac{x}{2} \) = \( \frac{x - 360}{2} \), so 100 + 180 = \( \frac{4}{3} \) x and x = 210 μC.
6. (a) q_A = 300 - 210 = 90 μC, q_C = 210 μC and the magnitude of q_B = 360 - 210 = 150 μC.
7. (b) Energy before = \( \frac{1}{2} \) × 3 × 10⁻⁶ × (100)² + \( \frac{1}{2} \) × 2 × 10⁻⁶ × (180)² = 0.015 + 0.0324 = 4.74 × 10⁻² J.
8. Energy after = \( \frac{(90)^{2}}{2 \times 3} \) + \( \frac{(150)^{2}}{2 \times 2} \) + \( \frac{(210)^{2}}{2 \times 2} \) = 1350 + 5625 + 11025 = 18000 μJ = 1.8 × 10⁻² J.

Teacher's Note:
a) Charge on an isolated section of conductors (a set of plates joined together) is conserved when the circuit is completed.
b) Energy decreases because charge flows through the wires and is dissipated as heat.

23. The distance between the two plates of a cathode ray oscilloscope is 1 cm and the potential difference between the two plates is 1200 volts. If a particle of charge +1.6 × 10⁻¹⁹ coulomb and energy 2 KeV enters at right angle to the field, find the maximum deflection in the direction of field if the plates are 1.5 cm long.

Answer: Maximum deflection = 3.375 mm
1. Electric field between the plates E = \( \frac{1200}{0.01} \) = 1.2 × 10⁵ V/m.
2. The kinetic energy = 2 keV = 2000 × 1.6 × 10⁻¹⁹ J, so m v² = 2 × 2000 e.
3. Time in the plates t = \( \frac{L}{v} \), and the deflection y = \( \frac{1}{2} \) \( \frac{qE}{m} \) \( \frac{L^{2}}{v^{2}} \) = \( \frac{qE L^{2}}{2 m v^{2}} \) = \( \frac{qE L^{2}}{4 \times KE} \).
4. y = \( \frac{1.6 \times 10^{-19} \times 1.2 \times 10^{5} \times (0.015)^{2}}{4 \times 2000 \times 1.6 \times 10^{-19}} \) = \( \frac{27}{8000} \) m.
5. y = 3.375 × 10⁻³ m = 3.375 mm.

Teacher's Note:
a) The motion in the plates is projectile-like: uniform speed along the plates and uniform acceleration across them.
b) The deflection does not depend on the mass or charge of the particle separately, only on the ratio qE/KE.

24. A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity σ₀ = 7.4 × 10⁻¹² Ω⁻¹ m-1 (conductivity σ₀ = \( \frac{1}{resistivity} \)). If the charge on the plate at the instant t = 0 is q = 8.85 μC, calculate the leakage current at t = 2 second.

Answer: Leakage current ≈ 1.06 μA
1. The capacitor C = \( \frac{K ε_0 A}{d} \) and the resistance of the dielectric R = \( \frac{d}{σ_0 A} \), so the time constant τ = RC = \( \frac{K ε_0}{σ_0} \).
2. τ = \( \frac{5 \times 8.85 \times 10^{-12}}{7.4 \times 10^{-12}} \) = 5.98 s.
3. The charge leaks as q = q₀ e^-t/τ, so the leakage current I = \( \frac{q_0}{τ} \) e^-t/τ.
4. I = \( \frac{8.85 \times 10^{-6}}{5.98} \) × e^-2/5.98 = 1.48 × 10⁻⁶ × 0.716.
5. I ≈ 1.06 × 10⁻⁶ A ≈ 1.1 μA.

Teacher's Note:
a) The time constant of a leaky capacitor, ε₀K/σ₀, does not depend on its dimensions.
b) The printed key of 0.24 μA does not follow from these data; the leakage current at t = 2 s is about 1.06 μA.

25. Find the potential difference V_A - V_B between points A and B of the circuit shown in figure

[Figure: A circuit with a top wire containing C₁ (left), a junction A and C₂ (right). The bottom wire contains cell ε₁ (left), a junction B and cell ε₂ (right). A capacitor C₃ joins A to B. In both cells the longer plate is on the right-hand side]

Answer: V_A - V_B = \( \frac{C_2 ε_2 - C_1 ε_1}{C_1 + C_2 + C_3} \)
1. Take V_B = 0. The cell ε₁ has its positive terminal on the B side, so the left end of the circuit is at potential -ε₁. The cell ε₂ has its positive terminal on the right, so the right end is at +ε₂.
2. The point A is connected only to capacitor plates, so the total charge on those three plates is zero: C₁(V_A + ε₁) + C₂(V_A - ε₂) + C₃(V_A - 0) = 0.
3. V_A(C₁ + C₂ + C₃) = C₂ε₂ - C₁ε₁.
4. V_A - V_B = \( \frac{C_2 ε_2 - C_1 ε_1}{C_1 + C_2 + C_3} \).

Teacher's Note:
a) No current flows in the steady state, so the charge on the isolated node A is conserved at zero.
b) The sign of the result depends on the polarities of the two cells.

26. A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half and a charge -Q is uniformly distributed along the lower half, as shown in the figure. Calculate the electric field E at P, the centre of the semicircle.

[Figure: A thin glass rod bent into a semicircle of radius r, opening to the right, with its centre P marked as a dot and O marked on the rod at its leftmost point. The end A is at the top and the end B is at the bottom. The upper half (from A to O) carries plus signs (charge +Q) and the lower half (from O to B) carries minus signs (charge -Q)]

Answer: E = \( \frac{Q}{π^{2} ε_0 r^{2}} \), directed from the positive half towards the negative half, parallel to AB
1. Each half of the semicircle is a quarter circle of length \( \frac{π r}{2} \), so the linear charge density λ = \( \frac{Q}{π r/2} \) = \( \frac{2Q}{π r} \) (positive on the upper half, negative on the lower half).
2. For an element at angle φ from the line OP, the field at P has magnitude \( \frac{k λ r dφ}{r^{2}} \). Resolve it parallel and perpendicular to OP.
3. The components along OP cancel between the upper and lower halves, since the upper half pushes away from itself and the lower half pulls towards itself, with equal magnitudes.
4. The components perpendicular to OP add. For one quarter circle the perpendicular component = \( \frac{kλ}{r} \) ∫ cos φ dφ over 0 to π/2 = \( \frac{kλ}{r} \).
5. Total E = 2 × \( \frac{kλ}{r} \) = \( \frac{2k}{r} \) × \( \frac{2Q}{π r} \) = \( \frac{4kQ}{π r^{2}} \) = \( \frac{Q}{π^{2} ε_0 r^{2}} \).
6. The field is directed parallel to the line AB, from the positively charged half towards the negatively charged half (from A towards B).

Teacher's Note:
a) Use the symmetry of the arrangement to see which components of the field cancel.
b) The printed key shows Q/(2π²ε₀R²), which is half of the value obtained from the integration above.
c) Check: a quarter circle of charge Q gives \( \frac{2\sqrt{2}}{π} \) kQ/r² at 45°; two of them give a vertical resultant of \( \frac{4}{π} \) kQ/r².

Please click on below link to download pdf file for CBSE Class 12 Physics Electrostatics Subjective Problems.