JEE Physics Electrostatic Potential and Capacitance MCQs Set 09

Answer: (a) 105

Explanation:
1. Capacitance \( C=\frac{K\varepsilon_0A}{d} \).
2. \( C'=\frac{3.5\,\varepsilon_0A}{2d}=\frac{3.5}{2}\times15=26.25\ \text{pF} \)
3. \( \frac{x}{4}=26.25\Rightarrow x=105 \).

Teacher's Note:
1. Capacitance is directly proportional to the dielectric constant and inversely proportional to the plate separation.
2. Both changes act on the same original value, so multiply the factors together.

Answer: (c) \( \frac{10}{3}\varepsilon_0 \) F

Explanation:
1. With a slab of thickness t and constant K: \( C=\frac{\varepsilon_0A}{d-t+t/K} \).
2. \( C=\frac{\varepsilon_0\times40\times10^{-4}}{2\times10^{-3}-1\times10^{-3}+0.2\times10^{-3}}=\frac{40\times10^{-4}}{1.2\times10^{-3}}\varepsilon_0=\frac{10}{3}\varepsilon_0 \) F.

Teacher's Note:
1. The slab behaves like an effective air gap of t/K, so the total effective separation is d - t + t/K.
2. Keep area in m² and thickness in m.

Answer: (a) 45°

Explanation:
1. Time in the field \( t=\frac{L}{v} \), transverse acceleration \( a=\frac{eE}{m} \).
2. \( \tan\theta=\frac{v_y}{v}=\frac{eEL}{mv^2}=\frac{eEL}{2K}=\frac{e(10)(0.1)}{2\times0.5\,e}=1 \) (K = 0.5 eV = 0.5e joule).
3. \( \theta=45^\circ \).

Teacher's Note:
1. The electron keeps its horizontal speed and picks up a vertical speed from the field.
2. Writing mv² as 2K lets you use the kinetic energy directly.

Answer: (a) 6 \( \mu \)F

Explanation:
1. Effective separation \( =d-\frac d2+\frac{d/2}{1.5}=\frac d2+\frac d3=\frac{5d}{6} \).
2. \( C'=\frac{6}{5}C=\frac65\times5=6\ \mu\text{F} \).

Teacher's Note:
1. Treat the slab as an air gap of thickness t/K.
2. A shorter effective gap always means a larger capacitance.

Answer: (d) \( \frac56 \)

Explanation:
1. Charges: \( Q_1=4\pi R^2\sigma,\ Q_2=4\pi(2R)^2\sigma=4Q_1 \), total \( 5Q_1 \).
2. After connection the potentials are equal, so charge is shared in proportion to radius: \( Q_2'=\frac23\times5Q_1=\frac{10}{3}Q_1 \).
3. \( \sigma'=\frac{Q_2'}{4\pi(2R)^2}=\frac{(10/3)(4\pi R^2\sigma)}{16\pi R^2}=\frac56\sigma \).

Teacher's Note:
1. Connected spheres reach a common potential, and V = kQ/R means Q is proportional to R.
2. Total charge is conserved.

Answer: (a) 225

Explanation:
1. Initial energy \( U_i=\frac12CV^2=\frac12(900\times10^{-6})(100)^2=4.5\ \text{J} \).
2. Charge is shared equally, so \( V'=50\ \text{V} \) and \( U_f=\frac12(2C)V'^2=2.25\ \text{J} \).
3. Loss \( =2.25\ \text{J}=225\times10^{-2}\ \text{J} \), so x = 225.

Teacher's Note:
1. When identical capacitors share charge, exactly half of the initial energy is lost.
2. Loss = (1/4) C V².

Answer: (a) 55 V

Explanation:
1. C1 stays at 100 V: \( C_1'=100\ \mu\text{F},\ Q_1=100\times10^{-6}\times100=10^{-2}\ \text{C} \).
2. C2 keeps its original charge: \( Q_2=10\times10^{-6}\times100=10^{-3}\ \text{C},\ C_2'=100\ \mu\text{F} \).
3. \( V=\frac{Q_1+Q_2}{C_1'+C_2'}=\frac{1.1\times10^{-2}}{2\times10^{-4}}=55\ \text{V} \).

Teacher's Note:
1. With the battery connected the voltage is fixed; with the battery removed the charge is fixed.
2. On joining in parallel, the total charge is shared at a common potential.

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

Explanation:
1. Charge on a conducting sphere \( Q=CV=4\pi\varepsilon_0RV \): it depends only on radius and potential, not on whether the sphere is solid or hollow.
2. Equal radii and equal potentials give equal charges, so the assertion is false.
3. The reason is a true statement (\( C=4\pi\varepsilon_0R \)).

Teacher's Note:
1. Charge on a conductor resides on its outer surface, so a hollow shell and a solid sphere of the same size hold the same charge at the same potential.
2. Capacitance of an isolated sphere depends only on its radius.

Answer: (b) \( \frac{\lambda}{2\varepsilon_0} \)

Explanation:
1. For a half ring of radius R: charge \( q=\lambda\pi R \) and every element is at distance R from the centre, so \( V=\frac{kq}{R}=\frac{\lambda\pi}{4\pi\varepsilon_0}=\frac{\lambda}{4\varepsilon_0} \) (independent of R).
2. Two half rings: \( V=\frac{\lambda}{4\varepsilon_0}+\frac{\lambda}{4\varepsilon_0}=\frac{\lambda}{2\varepsilon_0} \).

Teacher's Note:
1. Potential is a scalar, so the contributions simply add.
2. Each half ring gives the same potential because charge and distance are both proportional to R.

Answer: (c) -600 mJ

Explanation:
1. Work done by a uniform field: \( W=q\vec E\cdot\vec d \) with \( \vec d=(0-1)\hat i+(0-2)\hat j=-\hat i-2\hat j \).
2. \( W=2\times10^{-2}\times30\times(-1)=-0.6\ \text{J}=-600\ \text{mJ} \).

Teacher's Note:
1. Only the displacement component along the field counts.
2. The charge moves against the field, so the work is negative.

Answer: (c) V is constant for r less than or equal to R and falls as 1/r for r greater than R

Explanation:
1. Inside a conductor the field is zero, so the potential is constant and equal to its surface value \( \frac{kQ}{R} \).
2. Outside, the sphere acts as a point charge: \( V=\frac{kQ}{r} \) (falls as 1/r).

Teacher's Note:
1. The potential is continuous at the surface; only the field jumps in the presence of surface charge layers.
2. A flat region inside followed by a 1/r tail is the standard graph.