JEE Physics Electrostatic Potential and Capacitance MCQs Set 22

Answer: (d) (1) and (4) only

Explanation:
1. With the source connected, V is fixed. Capacitance becomes \( KC \).
2. Energy \( \tfrac12KCV^2 \) is K times (1 true) and charge \( KCV \) is K times (4 true).
3. Field \( V/d \) stays the same (2 false) and the force \( \tfrac12QE \) grows K times, not \( K^2 \) (3 false).

Teacher's Note:
1. Voltage is constant while the source remains connected.
2. Force between plates is half of charge times field.

Answer: (c) \( 3RC \)

Explanation:
1. The capacitor charges through R in series with the parallel combination of 6R and 3R (battery treated as ideal).
2. Parallel combination \( \dfrac{6R\times3R}{9R}=2R \), so the total resistance is \( R+2R=3R \).
3. Time constant \( \tau=3RC \).

Teacher's Note:
1. Remove the source and find the resistance seen by the capacitor.
2. An ideal battery acts as a short circuit when finding the time constant.

Answer: (a) 10

Explanation:
1. Losing 63% of charge means one time constant: \( \tau=RC=5 \) ms.
2. With 2R the time constant doubles: \( \tau'=2RC=10 \) ms, so K = 10.

Teacher's Note:
1. 63% is \( 1-e^{-1} \), i.e. one time constant.
2. Time constant is proportional to R.

Answer: (a) zero

Explanation:
1. Voltage on the 2 microfarad capacitor: \( \dfrac{20}{2}=10 \) V. Voltage on the 5 microfarad capacitor: \( \dfrac{50}{5}=10 \) V.
2. The voltages are equal and the polarities agree, so no charge flows when S is closed and no heat is produced.

Teacher's Note:
1. Charge flows only if there is a potential difference between the joined plates.
2. Compute the voltage on each capacitor first.

Answer: (c) \( \dfrac{Qq}{Q-q} \)

Explanation:
1. At each contact the total charge divides in the fixed ratio of the capacitances, so a conductor holding x receives a share \( \dfrac{q}{Q}(x+Q) \) of the total (since the first contact gives \( q=\dfrac{C}{C+C'}Q \)).
2. In the limit the charge no longer changes: \( x=\dfrac qQ(x+Q) \), giving \( x(Q-q)=qQ \).
3. So \( x_{max}=\dfrac{Qq}{Q-q} \).

Teacher's Note:
1. The maximum is reached when contact no longer changes the conductor's charge.
2. Charges share in the ratio of capacitances.

Answer: (b) \( \dfrac{4\varepsilon_0AV_0}{3d} \)

Explanation:
1. Take plate 4 at 0 and plates 1 and 3 at \( V_0 \). Plates 2 and 5 are at a common potential \( V_b \): the series of \( 2C_0 \) and \( C_0 \) gives \( V_b=\dfrac{2V_0}{3} \).
2. Plate 3 faces plate 2 and plate 4: \( q_3=C_0(V_0-V_b)+C_0V_0=C_0V_0\left(\dfrac13+1\right)=\dfrac43C_0V_0 \).

Teacher's Note:
1. Find the potential of the middle group of plates first.
2. A plate's charge is the sum over its two faces.

Answer: (a) \( a=\dfrac be \)

Explanation:
1. Field at radius r: \( E=\dfrac{\lambda}{2\pi\varepsilon_0r} \); the maximum is at \( r=a \): \( E_b=\dfrac{\lambda}{2\pi\varepsilon_0a} \).
2. Potential difference \( V=\int_a^bE\,dr=E_ba\ln\dfrac ba \).
3. Maximise: \( \dfrac{dV}{da}=E_b\left(\ln\dfrac ba-1\right)=0 \), so \( a=\dfrac be \).

Teacher's Note:
1. Express the voltage in terms of the breakdown field and the inner radius.
2. Set the derivative to zero.

Answer: (d) 75 V

Explanation:
1. Fields in the layers are in the ratio 1 : 3, so the potential drops (equal thicknesses) are also in the ratio 1 : 3.
2. Drop in glass \( =\dfrac14\times300=75 \) V (and 225 V in paraffin).

Teacher's Note:
1. Equal thicknesses mean voltage drops are proportional to the fields.
2. The two drops add up to 300 V.

Answer: (a) \( 8.0\times10^{-7} \) C/m²

Explanation:
1. The free charge density equals the displacement \( \sigma=\varepsilon_0K_gE_g \).
2. Value: \( 8.85\times10^{-12}\times6\times1.5\times10^4=7.97\times10^{-7} \) C/m².

Teacher's Note:
1. The free surface charge density equals the electric displacement D.
2. Use either layer: \( K E \) is the same.

Answer: (c) \( \dfrac{\varepsilon_0AV_0^2(K-1)}{2dK} \)

Explanation:
1. Charge stays fixed. Initial energy \( U_i=\dfrac12\dfrac{\varepsilon_0A}{d}V_0^2 \); final energy \( U_f=\dfrac{U_i}{K} \).
2. The slab is pulled in, so the system does work \( U_i-U_f=\dfrac{\varepsilon_0AV_0^2}{2d}\left(1-\dfrac1K\right)=\dfrac{\varepsilon_0AV_0^2(K-1)}{2dK} \).

Teacher's Note:
1. With the battery disconnected, the charge is constant and the energy falls by the factor K.
2. The lost electrical energy is work done on the man.

Answer: (d) 12 V

Explanation:
1. Let the empty capacitor have capacitance \( C_0 \). Its initial charge is \( 156C_0 \).
2. The non-linear capacitor holds charge \( KC_0V=\alpha C_0V^2 \) at voltage V.
3. Charge conservation: \( 156C_0=C_0V+C_0V^2 \), so \( V^2+V-156=0 \) giving \( V=12 \) V.

Teacher's Note:
1. Charge is conserved when the capacitors are joined.
2. The charge on the non-linear capacitor is not proportional to V.

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

Explanation:
1. Total capacitance \( 3C \). The battery does work \( QV=3CV^2 \).
2. Energy stored \( \tfrac12(3C)V^2=\tfrac32CV^2 \).
3. Heat \( =3CV^2-\tfrac32CV^2=\tfrac32CV^2 \).

Teacher's Note:
1. Half of the work done by the battery is stored and half is lost as heat.
2. Heat in charging does not depend on the resistance.

Answer: (a) \( -\dfrac12(k-1)CV^2 \)

Explanation:
1. Capacitance rises by \( (k-1)C \) and the battery supplies charge \( (k-1)CV \), doing work \( (k-1)CV^2 \).
2. Stored energy rises by \( \tfrac12(k-1)CV^2 \).
3. Energy balance: \( W_{agent}+(k-1)CV^2=\tfrac12(k-1)CV^2 \), so \( W_{agent}=-\tfrac12(k-1)CV^2 \) (the slab is pulled in).

Teacher's Note:
1. Include the battery work in the energy balance.
2. The negative sign means the slab is drawn inward.

Answer: (b) \( \dfrac{(k+2)(k-1)CV^2}{6} \)

Explanation:
1. Total charge \( Q=(k+2)CV \). With the slab: \( C_{tot}=(k+2)C \) and \( U_1=\dfrac{Q^2}{2(k+2)C}=\dfrac{(k+2)CV^2}{2} \).
2. After removal: \( C_{tot}=3C \) and \( U_2=\dfrac{Q^2}{6C}=\dfrac{(k+2)^2CV^2}{6} \).
3. Work \( =U_2-U_1=\dfrac{(k+2)CV^2}{6}\left[(k+2)-3\right]=\dfrac{(k+2)(k-1)CV^2}{6} \).

Teacher's Note:
1. With the battery disconnected, the total charge is fixed.
2. Work by the agent equals the increase in energy.

Answer: (c) \( \dfrac{\pi\varepsilon_0K_1S}{2d} \)

Explanation:
1. Thin layers are in series: \( \dfrac1C=\int_0^d\dfrac{dx}{\varepsilon_0K S}=\dfrac{1}{\varepsilon_0K_1S}\int_0^d\dfrac{dx}{1+\sin(\pi x/d)} \).
2. With \( \theta=\pi x/d \): \( \int_0^d\dfrac{dx}{1+\sin(\pi x/d)}=\dfrac d\pi\int_0^\pi\dfrac{d\theta}{1+\sin\theta}=\dfrac{2d}{\pi} \).
3. So \( \dfrac1C=\dfrac{2d}{\pi\varepsilon_0K_1S} \) and \( C=\dfrac{\pi\varepsilon_0K_1S}{2d} \).

Teacher's Note:
1. Layers of different dielectric constant stacked across the gap are in series.
2. Evaluate \( \int_0^\pi\dfrac{d\theta}{1+\sin\theta}=2 \).

Answer: (d) 13.86 s

Explanation:
1. Time constant \( \tau=RC=2.5\times10^6\times4\times10^{-6}=10 \) s.
2. During charging \( V_C=12(1-e^{-t/\tau}) \) and \( V_R=12e^{-t/\tau} \).
3. Setting \( V_C=3V_R \): \( 1-x=3x \) so \( e^{-t/\tau}=\tfrac14 \) and \( t=\tau\ln4=10\times2\times0.693=13.86 \) s.

Teacher's Note:
1. Capacitor and resistor voltages add up to the emf.
2. Solve for the exponential factor first.

Answer: (d) +48 microcoulomb

Explanation:
1. The 4 microfarad capacitor takes +80 microcoulomb on its upper plate and -80 on its lower plate. The joined upper plates of the 2 and 3 microfarad capacitors (initially neutral) therefore carry a total of +80 microcoulomb.
2. They are in parallel with the earth connection, so they share the charge in the ratio of capacitances 2 : 3: \( q_3=80\times\dfrac35=48 \) microcoulomb.

Teacher's Note:
1. Charge on an isolated group of plates is conserved.
2. Parallel capacitors at the same voltage share charge in proportion to capacitance.

Answer: (b) 420 N/C

Explanation:
1. The branch: 4 microfarad in series with 12 microfarad (3 + 9) gives 3 microfarad, so its charge is \( 3\times8=24 \) microcoulomb (the same on the 4 microfarad capacitor).
2. Voltage across the parallel pair \( =24/12=2 \) V, so the 9 microfarad capacitor holds \( 9\times2=18 \) microcoulomb.
3. Q = 24 + 18 = 42 microcoulomb and \( E=\dfrac{kQ}{r^2}=\dfrac{9\times10^9\times42\times10^{-6}}{900}=420 \) N/C.

Teacher's Note:
1. Series capacitors carry the same charge.
2. Reduce the parallel pair first.

Answer: (a) \( \dfrac13\left(\dfrac12CV_0^2\right) \)

Explanation:
1. Each step raises the source voltage by \( \Delta V=V_0/3 \). Energy dissipated in a step is \( \tfrac12C(\Delta V)^2=\tfrac12C\dfrac{V_0^2}{9} \).
2. Three equal steps: \( 3\times\tfrac12C\dfrac{V_0^2}9=\dfrac13\left(\tfrac12CV_0^2\right) \).

Teacher's Note:
1. Charging through a resistor by a voltage step wastes half the work, i.e. \( \tfrac12C(\Delta V)^2 \).
2. Add the three steps.

Answer: (c) \( E_C=E_D \)

Explanation:
1. Stored energy \( E_C=\tfrac12CV_0^2 \).
2. The source supplies \( QV_0=CV_0^2 \), so the heat is \( E_D=CV_0^2-\tfrac12CV_0^2=\tfrac12CV_0^2 \).
3. Hence \( E_C=E_D \).

Teacher's Note:
1. Half the energy from the source goes to the capacitor and half is dissipated.
2. This result is independent of R.

Answer: (c) 1.5

Explanation:
1. C3 initially holds \( Q=1.0\times8=8 \) microcoulomb. After connection it holds 5 microcoulomb, so the series branch takes 3 microcoulomb at the common voltage \( V=5 \) V.
2. Equivalent capacitance of the branch \( =\dfrac3 5=0.6 \) microfarad.
3. Series of \( \varepsilon_r\times1 \) and 1: \( \dfrac{\varepsilon_r}{\varepsilon_r+1}=0.6 \), so \( \varepsilon_r=1.5 \).

Teacher's Note:
1. Charge is conserved when C3 is connected across the series branch.
2. The common voltage is that across C3.