JEE Physics Electrostatic Potential and Capacitance MCQs Set 32

Answer: (c) 8.5

Explanation:
1. Minimum separation \( d=\dfrac V{E_{max}}=\dfrac{500}{10^6}=5\times10^{-4} m \).
2. \( C=\dfrac{K\epsilon_0A}d\Rightarrow K=\dfrac{15\times10^{-12}\times5\times10^{-4}}{8.86\times10^{-12}\times10^{-4}}=8.5 \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (a) 508 pJ

Explanation:
1. Charge is constant: \( U_f=\dfrac{U_i}K \), \( U_i=\tfrac12CV^2=600 \) pJ.
2. Work done by the capacitor \( =U_i-U_f=600\left(1-\dfrac1{6.5}\right)=508 \) pJ.

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (c) 32

Explanation:
1. Each series string must have at least \( \dfrac{1000}{300}\to4 \) capacitors: \( \dfrac14\ \mu F \) per string.
2. For 2 \( \mu F \) we need \( \dfrac2{1/4}=8 \) strings in parallel: \( 4\times8=32 \) capacitors.

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

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

Explanation:
1. Free charge density \( \sigma=K\epsilon_0E \).
2. \( \sigma=2.2\times8.85\times10^{-12}\times3\times10^4=5.8\times10^{-7}\approx6\times10^{-7} C/m ^2 \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (c) \( -\dfrac{2\sigma}{\epsilon_0}\hat k \)

Explanation:
1. Each sheet gives \( \dfrac{\sigma}{2\epsilon_0} \) directed away from positive and towards negative sheets. At \( P \): the positive sheet above pushes down (\( \sigma \)), the \( -2\sigma \) sheet below attracts down (\( 2\sigma \)), the \( -\sigma \) sheet below attracts down (\( \sigma \)).
2. Total \( =\dfrac{(1+2+1)\sigma}{2\epsilon_0}=\dfrac{2\sigma}{\epsilon_0} \) downward, \( -\dfrac{2\sigma}{\epsilon_0}\hat k \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (d) \( \dfrac14C(V_1-V_2)^2 \)

Explanation:
1. Common potential \( V=\dfrac{V_1+V_2}2 \). Initial energy \( \tfrac12C(V_1^2+V_2^2) \), final \( \tfrac12(2C)V^2=\dfrac C4(V_1+V_2)^2 \).
2. Loss \( =\dfrac C4\left[2V_1^2+2V_2^2-(V_1+V_2)^2\right]=\dfrac C4(V_1-V_2)^2 \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (b) \( \dfrac1K=\dfrac1{K_1+K_2}+\dfrac1{2K_3} \)

Explanation:
1. Upper half (parallel): \( C_u=\dfrac{\epsilon_0(K_1+K_2)A}{2\cdot d/2}=\dfrac{\epsilon_0(K_1+K_2)A}d \). Lower half: \( C_l=\dfrac{\epsilon_0K_3A}{d/2}=\dfrac{2\epsilon_0K_3A}d \).
2. Series: \( \dfrac1C=\dfrac d{\epsilon_0A}\left[\dfrac1{K_1+K_2}+\dfrac1{2K_3}\right] \), so \( \dfrac1K=\dfrac1{K_1+K_2}+\dfrac1{2K_3} \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (d) zero

Explanation:
1. The source has no internal resistance and there is no resistor in series with the capacitor, so the time constant of the charging branch is zero.
2. The capacitor charges essentially instantaneously to the full \( 1.5 V \); the time to reach \( 0.75 V \) is zero.

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (b) \( \dfrac32CV^2 \)

Explanation:
1. Charges: \( CV \) and \( 2C\cdot2V=4CV \). Opposite connection: net charge \( =3CV \) on total capacitance \( 3C \), so \( V'=V \).
2. \( U=\tfrac12(3C)V^2=\dfrac32CV^2 \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (d) (b) and (d) only

Explanation:
1. The battery is disconnected, so the charge \( Q \) is constant (\( a \) is false). \( C=\dfrac{\epsilon_0A}d \) decreases (\( c \) is false).
2. \( V=\dfrac QC \) increases (\( b \) true) and \( U=\dfrac{Q^2}{2C} \) increases (\( d \) true), the extra energy coming from the work done by the person.

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (b) 1.5

Explanation:
1. First, \( C_3 \) holds \( Q=1\times8=8\ \mu C \). After sharing, the voltage \( V \) is common to \( C_3 \) and the series pair \( \dfrac{\epsilon_r}{\epsilon_r+1}\ \mu F \).
2. \( 5=1\times V \), so \( V=5 V \) and the total charge \( 8=5\left(1+\dfrac{\epsilon_r}{\epsilon_r+1}\right)\Rightarrow\dfrac{\epsilon_r}{\epsilon_r+1}=\dfrac35\Rightarrow\epsilon_r=1.5 \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.

Answer: (c) \( \dfrac{3V}{K+2} \)

Explanation:
1. Charge conserved: \( Q=3CV \). New capacitance \( KC+2C \).
2. \( V'=\dfrac{3CV}{(K+2)C}=\dfrac{3V}{K+2} \).

Teacher's Note:
1. For capacitors use \( C = K epsilon_0 A / d \), add in parallel and add reciprocals in series, and decide first whether the battery stays connected (\( V \) fixed) or is removed (\( Q \) fixed).
2. Energy stored is \( Q^2/2C = CV^2/2 \); compare before and after to get the work done.