CBSE Class 12 Physics Electrostatic Potential and Capacitance MCQs Set 06

Answer: (a) 1 × 10⁻⁴ J

Explanation:
1. W = q(V_B − V_A) = 2 × 10⁻⁶ × 50
2. W = 1 × 10⁻⁴ J

Teacher's Note:
1. Potential difference is work per unit charge, so W = qΔV.
2. Convert μC to C before multiplying.

Answer: (c) 150 V

Explanation:
1. V = k \( \frac{q}{r} \) = \( \frac{9 \times 10^9 \times 5 \times 10^{-9}}{0.3} \)
2. V = \( \frac{45}{0.3} \) = 150 V

Teacher's Note:
1. Potential uses r, not r²; 500 V comes from wrongly using r².
2. 1 nC = 10⁻⁹ C cancels neatly with k = 9 × 10⁹.

Answer: (b) V is zero but E is not zero

Explanation:
1. The potentials kq/r and −kq/r add algebraically to give V = 0.
2. The fields of +q and −q at the midpoint both point towards −q, so they add and E ≠ 0.
3. So V is zero but E is not zero.

Teacher's Note:
1. V is a scalar and adds with signs; E is a vector and adds with directions.
2. Zero potential at a point does not mean zero field there.

Answer: (d) zero

Explanation:
1. V = \( \frac{1}{4\pi\varepsilon_0} \frac{p \cos\theta}{r^2} \)
2. On the equatorial line θ = 90°, so cos θ = 0 and V = 0.

Teacher's Note:
1. Every point on the equatorial line is equidistant from +q and −q.
2. The field on the equatorial line is not zero; it points opposite to p.

Answer: (a) 90°

Explanation:
1. No work is done in moving a charge along an equipotential surface, so the force (and field) has no component along it.
2. Hence the field is always perpendicular (90°) to the equipotential surface.

Teacher's Note:
1. This is why field lines meet a conductor's surface at right angles.
2. If E had a component along the surface, V would change along it.

Answer: (c) 8 V m⁻¹ along −x

Explanation:
1. E = \( -\frac{dV}{dx} \) = −8x
2. At x = 1 m, E = −8 V m⁻¹, i.e. 8 V m⁻¹ along −x.

Teacher's Note:
1. The minus sign means E points towards decreasing V.
2. Differentiate first, then substitute x.

Answer: (b) −0.9 J

Explanation:
1. U = k \( \frac{q_1 q_2}{r} \) = \( \frac{9 \times 10^9 \times 2 \times 10^{-6} \times (-3 \times 10^{-6})}{0.06} \)
2. U = \( \frac{-0.054}{0.06} \) = −0.9 J

Teacher's Note:
1. Keep the signs of the charges in U; a negative U means the charges attract.
2. 6 cm = 0.06 m.

Answer: (d) 4 × 10⁻⁴ J

Explanation:
1. W = pE(cos θ₁ − cos θ₂) = pE(cos 0° − cos 180°) = 2pE
2. W = 2 × 2 × 10⁻⁸ × 1 × 10⁴ = 4 × 10⁻⁴ J

Teacher's Note:
1. The energy goes from −pE to +pE, a change of 2pE.
2. 2 × 10⁻⁴ J would be the work to turn it only through 90°.

Answer: (a) E = 0 and V is the same as on the surface

Explanation:
1. The field inside a conductor (and its cavity) is zero.
2. E = −dV/dr = 0 means V does not change, so V everywhere inside equals the surface value kq/R.
3. Answer: E = 0 and V is the same as on the surface

Teacher's Note:
1. Zero field means constant potential, not zero potential.
2. The whole conductor is one equipotential volume.

Answer: (c) 177 pF

Explanation:
1. C = \( \frac{\varepsilon_0 A}{d} \) = \( \frac{8.854 \times 10^{-12} \times 0.02}{1 \times 10^{-3}} \)
2. C = 1.77 × 10⁻¹⁰ F = 177 pF

Teacher's Note:
1. 1 pF = 10⁻¹² F, so 1.77 × 10⁻¹⁰ F = 177 pF.
2. The farad is a very large unit; real capacitors are in μF, nF or pF.

Answer: (b) becomes 1/K times

Explanation:
1. After disconnection the charge Q stays constant.
2. C becomes KC, and U = \( \frac{Q^2}{2C} \), so U becomes 1/K times.

Teacher's Note:
1. With Q fixed use U = Q²/2C; with V fixed use U = ½CV².
2. The slab is pulled into the capacitor, so the stored energy decreases.

Answer: (d) 8 V

Explanation:
1. C_s = \( \frac{3 \times 6}{3 + 6} \) = 2 μF, so Q = C_sV = 2 × 12 = 24 μC on each capacitor.
2. V₃ = \( \frac{Q}{C} \) = \( \frac{24}{3} \) = 8 V (and 4 V across 6 μF)

Teacher's Note:
1. In series, the smaller capacitor takes the larger voltage.
2. Check: 8 V + 4 V = 12 V.

Answer: (a) 0.2 J

Explanation:
1. U = \( \frac{1}{2} CV^2 \) = \( \frac{1}{2} \) × 10 × 10⁻⁶ × (200)²
2. U = 5 × 10⁻⁶ × 4 × 10⁴ = 0.2 J

Teacher's Note:
1. Do not forget the factor ½; 0.4 J is the result without it.
2. Square the voltage before multiplying.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. C = Q/V, so its unit is C V⁻¹, called the farad. The Assertion is correct.
2. Air breaks down at about 3 × 10⁶ V m⁻¹, so the Reason is also correct.
3. Dielectric strength has nothing to do with the name of the unit of capacitance, so the Reason does not explain the Assertion.

Teacher's Note:
1. When both statements are true but unrelated, choose (b).
2. 1 F = 1 C V⁻¹.

Answer: (c) A is true, but R is false.

Explanation:
1. Free charges rearrange until the field inside vanishes, so the Assertion is true.
2. E = 0 inside means V is constant, equal to the surface value, which is generally not zero. The Reason is false.

Teacher's Note:
1. Constant potential and zero potential are different ideas.
2. A charged sphere has V = kq/R everywhere inside it.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. With the dielectric, E becomes E₀/K and V becomes V₀/K for the same Q.
2. C = Q/V therefore becomes K times larger. Both statements are true.
3. The fall in V for the same Q is exactly why C rises, so the Reason explains the Assertion.

Teacher's Note:
1. C = Kε₀A/d for a completely filled capacitor.
2. The dielectric's bound surface charges oppose the applied field.

Answer: (c) A is true, but R is false.

Explanation:
1. 1/C_s = 1/C₁ + 1/C₂ + … is larger than any single 1/C, so C_s is smaller than every C. The Assertion is true.
2. In series the capacitors carry the same charge, not the same potential difference. The Reason is false.

Teacher's Note:
1. Series: same Q. Parallel: same V.
2. In series the voltage divides in inverse proportion to C.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. At the midpoint of a dipole, V = 0 but E ≠ 0, so the Assertion is true.
2. E = −dV/dr depends on how fast V changes, so the Reason is true.
3. The Reason explains why a zero value of V can still go with a non-zero field.

Teacher's Note:
1. Likewise E = 0 can go with a non-zero (constant) V.
2. Think of slope versus height on a graph of V against r.

Answer: (d) A is false, but R is true.

Explanation:
1. Charge flows until both reach a common potential, and some energy is lost as heat in the connecting wires. The Assertion is false.
2. Charge is conserved during the flow, so the Reason is true.

Teacher's Note:
1. Charge is conserved but electrostatic energy is not.
2. The loss is C₁C₂(V₁ − V₂)²/2(C₁ + C₂).

Answer: (b) 354 pF

Explanation:
1. A = 0.2 × 0.1 = 0.02 m², d = 5 × 10⁻⁴ m
2. C = \( \frac{8.854 \times 10^{-12} \times 0.02}{5 \times 10^{-4}} \) = 3.54 × 10⁻¹⁰ F = 354 pF

Teacher's Note:
1. Convert cm² to m² and mm to m first.
2. Halving the gap would double C.

Answer: (c) 35.4 nC

Explanation:
1. Q = CV = 3.54 × 10⁻¹⁰ × 100
2. Q = 3.54 × 10⁻⁸ C = 35.4 nC

Teacher's Note:
1. One plate carries +Q and the other −Q.
2. 1 nC = 10⁻⁹ C.

Answer: (a) 177 nC

Explanation:
1. With the supply connected, V stays 100 V while C becomes 5 times larger.
2. Q = 5 × 35.4 nC = 177 nC

Teacher's Note:
1. Extra charge flows from the supply to keep V fixed.
2. 7.08 nC comes from wrongly dividing by K instead of multiplying.

Answer: (d) 1.77 × 10⁻⁶ J

Explanation:
1. U = \( \frac{1}{2} CV^2 \) = \( \frac{1}{2} \) × 3.54 × 10⁻¹⁰ × (100)²
2. U = 1.77 × 10⁻⁶ J

Teacher's Note:
1. 3.54 × 10⁻⁶ J is the value without the ½.
2. This energy is stored in the field between the plates.

Answer: (a) 2 × 10⁵ V m⁻¹

Explanation:
1. The field between the plates is uniform: E = \( \frac{V}{d} \)
2. E = \( \frac{100}{5 \times 10^{-4}} \) = 2 × 10⁵ V m⁻¹

Teacher's Note:
1. This is well below the breakdown field of air (about 3 × 10⁶ V m⁻¹).
2. Use d in metres.

Answer: (c) 6.75 J

Explanation:
1. U = \( \frac{1}{2} CV^2 \) = \( \frac{1}{2} \) × 150 × 10⁻⁶ × (300)²
2. U = 75 × 10⁻⁶ × 9 × 10⁴ = 6.75 J

Teacher's Note:
1. 13.5 J is the value without the ½.
2. A few joules released in about a millisecond gives a very bright flash.

Answer: (b) 45 mC

Explanation:
1. Q = CV = 150 × 10⁻⁶ × 300
2. Q = 0.045 C = 45 mC

Teacher's Note:
1. 1 mC = 10⁻³ C.
2. Q is directly proportional to V for a given capacitor.

Answer: (d) 1.69 J

Explanation:
1. U ∝ V², so halving V makes U one-fourth.
2. U = \( \frac{6.75}{4} \) ≈ 1.69 J

Teacher's Note:
1. 3.38 J is the trap from thinking U ∝ V.
2. Use ratios instead of recalculating from scratch.

Answer: (a) 300 μF

Explanation:
1. Two identical capacitors C in series give C_s = C/2.
2. C/2 = 150 μF, so C = 300 μF

Teacher's Note:
1. Series combination always lowers the capacitance.
2. 75 μF would be right for two capacitors in parallel.

Answer: (c) 150 V

Explanation:
1. V = \( \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} \) = \( \frac{150 \times 300 + 0}{150 + 150} \)
2. V = 150 V

Teacher's Note:
1. Charge is shared but conserved: 45 mC is now spread over 300 μF.
2. Half of the original energy is lost as heat in this process.