JEE Physics Electrostatic Potential and Capacitance MCQs Set 13

Answer: (c) \( \frac{k_1k_2(d_1+d_2)}{k_1d_2+k_2d_1} \)

Explanation:
1. Layers in series: \( \frac{d}{k}=\frac{d_1}{k_1}+\frac{d_2}{k_2}=\frac{k_2d_1+k_1d_2}{k_1k_2} \).
2. \( k=\frac{k_1k_2(d_1+d_2)}{k_1d_2+k_2d_1} \)

Teacher's Note:
1. Capacitances in series add as reciprocals, so d/k adds for layers placed one after another.

Answer: (a) \( 6\times10^{-7}\ \text{C/m}^2 \)

Explanation:
1. Free charge density on the plate \( \sigma=K\varepsilon_0E \).
2. \( \sigma=2.2\times8.85\times10^{-12}\times3\times10^4\approx5.8\times10^{-7}\approx6\times10^{-7}\ \text{C/m}^2 \)

Teacher's Note:
1. The field in a dielectric is E = σ/(Kε0), so σ = Kε0E.

Answer: (b) \( \frac ab \)

Explanation:
1. Inner charged, outer earthed: \( C_1=\frac{4\pi\varepsilon_0ab}{b-a} \).
2. Outer charged, inner earthed: the outer surface also holds charge facing the surroundings, and the capacity is \( C_2=4\pi\varepsilon_0b+\frac{4\pi\varepsilon_0ab}{b-a}=\frac{4\pi\varepsilon_0b^2}{b-a} \).
3. \( \frac{C_1}{C_2}=\frac ab \)

Teacher's Note:
1. When the outer sphere is charged, the capacity includes the outer sphere's own capacitance to the surroundings in parallel with the shell-to-shell capacitance.

Answer: (b) 500 V/m

Explanation:
1. \( E=\frac Vd=\frac{10}{0.02}=500\ \text{V/m} \)

Teacher's Note:
1. Convert the separation to metres before dividing.

Answer: (a) 3/2

Explanation:
1. \( C_1+C_2=\frac{25}{6}\cdot\frac{C_1C_2}{C_1+C_2}\Rightarrow\frac{(C_1+C_2)^2}{C_1C_2}=\frac{25}{6} \)
2. With \( x=C_2/C_1 \): \( x+\frac1x+2=\frac{25}{6}\Rightarrow6x^2-13x+6=0 \).
3. Roots x = 3/2 or 2/3; since C2 is greater, x = 3/2.

Teacher's Note:
1. Express both combinations in terms of the ratio x and solve the quadratic.

Answer: (a) \( 2.65\times10^6\ \text{m/s} \)

Explanation:
1. The electron gains kinetic energy \( eV=20\ \text{eV} \).
2. \( v=\sqrt{\frac{2eV}{m}}=\sqrt{\frac{2(1.6\times10^{-19})(20)}{9.11\times10^{-31}}}=2.65\times10^6\ \text{m/s} \)

Teacher's Note:
1. Energy conservation: kinetic energy gained equals charge times potential difference.

Answer: (c) Remains constant

Explanation:
1. The battery holds the potential difference at V.
2. \( E=\frac Vd \) depends only on V and d, so it stays the same.

Teacher's Note:
1. With the battery connected, V and d are fixed so E is unchanged; only the charge increases.

Answer: (b) 4 V

Explanation:
1. Charges: \( Q_1=36\ \mu\text{C},\ Q_2=72\ \mu\text{C} \) in opposite sense, so net \( 72-36=36\ \mu\text{C} \) is shared in parallel.
2. Common potential \( V=\frac{36}{3+6}=4\ \text{V} \).

Teacher's Note:
1. Joining opposite plates cancels charge; the remainder redistributes in parallel.

Answer: (a) 2

Explanation:
1. \( C'=\frac{\varepsilon_0A}{(d-t)+t/K}=\frac{\varepsilon_0A}{\frac d2+\frac{d}{2K}}=\frac43\cdot\frac{\varepsilon_0A}{d} \)
2. \( \frac12+\frac1{2K}=\frac34\Rightarrow K=2 \)

Teacher's Note:
1. A slab of thickness t and constant K reduces the effective gap to d − t + t/K.

Answer: (d) Two in series and one in parallel

Explanation:
1. Two in series give 2 μF; adding one 4 μF in parallel gives 6 μF.
2. Check the others: all series 4/3 μF, all parallel 12 μF, two parallel then series 8/3 μF.

Teacher's Note:
1. Parallel adds capacitance, series reduces it; try combinations until the sum is 6 μF.

Answer: (b) 3W/4

Explanation:
1. \( W=\frac12C(40^2-20^2)=600C \)
2. \( W'=\frac12C(50^2-40^2)=450C=\frac34W \)

Teacher's Note:
1. Work equals the change in stored energy, proportional to the change in V².

Answer: (a) \( \frac\pi4\sqrt{LC} \)

Explanation:
1. \( q=q_0\cos\omega t,\ \omega=\frac1{\sqrt{LC}} \)
2. Electric energy is half of total when \( \cos^2\omega t=\tfrac12\Rightarrow\omega t=\tfrac\pi4 \), so \( t=\frac\pi4\sqrt{LC} \)

Teacher's Note:
1. Energy oscillates between the capacitor and the inductor; equal sharing first occurs at one-eighth of the period.

Answer: (b) 10 ms

Explanation:
1. \( i=i_0e^{-t/RC} \)
2. \( e^{-t/RC}=e^{-2}\Rightarrow t=2RC=2(1000)(5\times10^{-6})=10\ \text{ms} \)

Teacher's Note:
1. The current falls by a factor e every RC seconds.

Answer: (c) 5

Explanation:
1. Charge C1V is shared in parallel: \( V_f=\frac{C_1V}{C_1+C_2}=\frac V6 \)
2. \( C_1+C_2=6C_1\Rightarrow\frac{C_2}{C_1}=5 \)

Teacher's Note:
1. Charge is conserved when a charged and an uncharged capacitor are joined in parallel.

Answer: (c) 7

Explanation:
1. \( i=i_0e^{-t/\tau}:\ t_1=\tau\ln8=3\tau\ln2,\ t_2=\tau\ln16=4\tau\ln2 \)
2. \( \frac{t_2+t_1}{t_2-t_1}=\frac{7}{1}=7 \)

Teacher's Note:
1. Take logarithms of the exponential decay to get each time in terms of the time constant.

Answer: (b) 2 mV

Explanation:
1. \( V=Ed=10\times0.2\times10^{-3}=2\times10^{-3}\ \text{V}=2\ \text{mV} \)

Teacher's Note:
1. Breakdown voltage is the maximum field times the plate separation.

Answer: (c) 400 μJ

Explanation:
1. Time constant \( RC=0.2\ \text{s} \), so \( t_1=0.7RC,\ t_2=1.4RC \). With \( e^{-0.7}=\tfrac12,\ e^{-1.4}=\tfrac14 \), charge fractions are 0.5 and 0.75.
2. \( \Delta U=\frac12C\varepsilon^2(0.75^2-0.5^2)=\frac12(10\times10^{-6})(256)(0.3125)=400\ \mu\text{J} \)

Teacher's Note:
1. Charge grows as 1 − e^(−t/RC), and the stored energy is q²/2C.

Answer: (a) 1

Explanation:
1. In series both capacitors carry the same charge Q.
2. Force between the plates \( F=\frac{Q^2}{2\varepsilon_0A} \) depends on Q and plate area only, so FA/FB = 1.

Teacher's Note:
1. The attractive force depends on the charge, not the separation, so equal charge gives equal force.

Answer: (c) 90

Explanation:
1. \( U_i=\frac12(3C)V^2=\frac32CV^2 \). Initial charges CV and 2CV.
2. After joining opposite plates, the charges q1 and q2 satisfy \( q_1/C=-q_2/2C \) (equal potentials of the joined plates) and charge conservation \( q_1-q_2=-CV \), giving \( q_1=-\frac{CV}{3},\ q_2=\frac{2CV}{3} \).
3. \( U_f=\frac{q_1^2}{2C}+\frac{q_2^2}{4C}=\frac{CV^2}{6} \), so \( 10\frac{U_i}{U_f}=10\times9=90 \)

Teacher's Note:
1. Use charge conservation on the joined plates and the zero-current condition (no net potential difference around the loop).

Answer: (c) 6 μF

Explanation:
1. \( C'=\frac{\varepsilon_0A}{\frac{d}{3K_1}+\frac{2d}{3K_2}}=\frac{\varepsilon_0A}{\frac d6+\frac d6}=\frac{3\varepsilon_0A}{d}=3C \) = 6 μF

Teacher's Note:
1. Slabs filling the gap in series add their reduced thicknesses t/K.

Answer: (c) 9

Explanation:
1. Series: \( C_s=\frac C3,\ U_S=\frac12\cdot\frac C3V^2 \). Parallel: \( C_p=3C,\ U_P=\frac12\cdot3CV^2 \).
2. \( \frac{U_P}{U_S}=\frac{3C}{C/3}=9 \)

Teacher's Note:
1. At fixed V the stored energy is proportional to the equivalent capacitance.