Question

Between the plates of a parallel plate capacitor of plate area A and capacity 0.025 \( \mu \)F, a metal plate of area A and thickness equal to \( \frac{1}{3} \) of the separation between the plates of the capacitor is introduced. If the capacitor is charged to 100 V, then the amount of work done to remove the metal plate from the capacitor is

• A. \( 62.5 \) \( \mu \)J
• B. \( 30.2 \) \( \mu \)J
• C. \( 52.6 \) \( \mu \)J
• D. \( 35.4 \) \( \mu \)J

Hint:

When a metal plate is inserted into a capacitor and then removed while the capacitor is connected to a voltage source, the work done to remove the plate is equal to the change in the energy stored in the capacitor. First, find the initial capacitance with the metal plate inserted (equivalent of two capacitors in series). Then, find the final capacitance after the metal plate is removed (the original capacitance). Calculate the initial and final energies stored in the capacitor using \( U = \frac{1}{2} C V^2 \) (assuming voltage remains constant). The work done to remove the plate is the difference between the final and initial energies.

Answer 1

The Correct Option is A

Let the initial separation between the plates of the capacitor be \( d \).
The thickness of the metal plate introduced is \( t = \frac{1}{3} d \).
When a metal plate is introduced between the plates of a capacitor, it effectively forms two capacitors in series.
The new separation between the plates and the metal plate is \( d - t = d - \frac{1}{3} d = \frac{2}{3} d \).
This gap is divided into two equal parts by the metal plate (assuming it's centrally placed, although the exact position doesn't affect the final capacitance).
So, each of the two new capacitors has a separation of \( \frac{1}{2} (\frac{2}{3} d) = \frac{1}{3} d \).
The initial capacitance of the parallel plate capacitor is \( C_0 = \frac{\epsilon_0 A}{d} = 0.
025 \times 10^{-6} \) F.
When the metal plate is introduced, the system can be considered as two capacitors in series, each with plate area \( A \) and separation \( \frac{1}{3} d \).
The capacitance of each of these capacitors is \( C' = \frac{\epsilon_0 A}{(d/3)} = 3 \frac{\epsilon_0 A}{d} = 3 C_0 \).
The equivalent capacitance \( C_{eq} \) of two capacitors in series is given by \( \frac{1}{C_{eq}} = \frac{1}{C'} + \frac{1}{C'} = \frac{2}{C'} \).
So, \( C_{eq} = \frac{C'}{2} = \frac{3 C_0}{2} = \frac{3}{2} (0.
025 \times 10^{-6}) = 0.
0375 \times 10^{-6} \) F.
The initial energy stored in the capacitor before the metal plate is removed (with the metal plate inserted) when charged to 100 V is: \( U_i = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} (0.
0375 \times 10^{-6}) (100)^2 = \frac{1}{2} (0.
0375 \times 10^{-6}) (10000) = 0.
1875 \times 10^{-3} \) J \( = 187.
5 \) \( \mu \)J.
The final energy stored in the capacitor after the metal plate is removed (which is the original capacitor) when charged to 100 V is: \( U_f = \frac{1}{2} C_0 V^2 = \frac{1}{2} (0.
025 \times 10^{-6}) (100)^2 = \frac{1}{2} (0.
025 \times 10^{-6}) (10000) = 0.
125 \times 10^{-3} \) J \( = 125 \) \( \mu \)J.
The work done to remove the metal plate is the difference in the final and initial energies stored in the capacitor: \( W = U_f - U_i = 125 \times 10^{-6} - 187.
5 \times 10^{-6} = -62.
5 \times 10^{-6} \) J \( = -62.
5 \) \( \mu \)J.
The work done *by* the system is negative, so the work done *to remove* the plate is positive.
Let's re-evaluate the energy change.
When the metal plate is removed at constant charge, the potential difference changes.
Let's consider the charge \( Q = C_{eq} V_i = (0.
0375 \times 10^{-6}) (100) = 3.
75 \times 10^{-6} \) C.
The final energy with the metal plate removed is \( U_f = \frac{Q^2}{2 C_0} = \frac{(3.
75 \times 10^{-6})^2}{2 (0.
025 \times 10^{-6})} = \frac{14.
0625 \times 10^{-12}}{0.
05 \times 10^{-6}} = 281.
25 \times 10^{-6} \) J \( = 281.
25 \) \( \mu \)J.
The initial energy with the metal plate inserted is \( U_i = \frac{Q^2}{2 C_{eq}} = \frac{(3.
75 \times 10^{-6})^2}{2 (0.
0375 \times 10^{-6})} = \frac{14.
0625 \times 10^{-12}}{0.
075 \times 10^{-6}} = 187.
5 \times 10^{-6} \) J \( = 187.
5 \) \( \mu \)J.
The work done to remove the plate is \( W = U_f - U_i = 281.
25 - 187.
5 = 93.
75 \) \( \mu \)J.
This does not match any option.
Let's consider the case where the voltage remains constant at 100 V.
The work done to remove the plate is the change in energy supplied by the battery.
The energy supplied by the battery is \( \Delta U + W_{done\_on\_battery} \).
Let's use the force method.
The force on the metal plate is due to the electric field.
The electric field in each gap is \( E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0} \).
The force on the plate is \( F = Q E / 2 \) (factor of 1/2 because the field due to the plate itself doesn't exert force on it).
\( F = \frac{Q^2}{2 A \epsilon_0} \).
The work done to move it by \( \frac{1}{3} d \) is \( W = F \times \frac{1}{3} d = \frac{Q^2 d}{6 A \epsilon_0} \).
We know \( C_0 = \frac{\epsilon_0 A}{d} \), so \( \frac{d}{A \epsilon_0} = \frac{1}{C_0} \).
\( W = \frac{Q^2}{6 C_0} \).
\( Q = C_{eq} V = \frac{3}{2} C_0 V \).
\( W = \frac{(\frac{3}{2} C_0 V)^2}{6 C_0} = \frac{\frac{9}{4} C_0^2 V^2}{6 C_0} = \frac{3}{8} C_0 V^2 \).
\( W = \frac{3}{8} (0.
025 \times 10^{-6}) (100)^2 = \frac{3}{8} (0.
025 \times 10^{-6}) (10000) = \frac{3}{8} (0.
25 \times 10^{-3}) = 0.
09375 \times 10^{-3} \) J \( = 93.
75 \) \( \mu \)J.
Still no match.
There's likely a simpler approach.
The force on the metal plate is related to the energy density of the electric field.
The electric field in the gaps is \( E = V / (d/3) = 3V/d \).
Energy density \( u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \epsilon_0 \frac{9V^2}{d^2} \).
Force \( F = u A = \frac{9}{2} \epsilon_0 \frac{A V^2}{d^2} \).
Work \( W = F \times \frac{1}{3} d = \frac{3}{2} \epsilon_0 \frac{A V^2}{d} = \frac{3}{2} C_0 V^2 \).
\( W = \frac{3}{2} (0.
025 \times 10^{-6}) (100)^2 = \frac{3}{2} (0.
25 \times 10^{-3}) = 0.
375 \times 10^{-3} \) J \( = 375 \) \( \mu \)J.
No match.
Let's consider the change in capacitance.
\( C_i = 1.
5 C_0 \).
\( C_f = C_0 \).
If charge is constant \( Q = C_i V = 1.
5 C_0 V \).
\( W = \Delta U = \frac{Q^2}{2 C_f} - \frac{Q^2}{2 C_i} = \frac{(1.
5 C_0 V)^2}{2 C_0} - \frac{(1.
5 C_0 V)^2}{3 C_0} = \frac{2.
25}{2} C_0 V^2 - \frac{2.
25}{3} C_0 V^2 = 1.
125 C_0 V^2 - 0.
75 C_0 V^2 = 0.
375 C_0 V^2 \).
If voltage is constant, \( W = \Delta U_{capacitor} - W_{battery} = \frac{1}{2} C_f V^2 - \frac{1}{2} C_i V^2 - V \Delta Q = \frac{1}{2} (C_0 - 1.
5 C_0) V^2 - V (C_0 - 1.
5 C_0) V = -0.
25 C_0 V^2 + 0.
5 C_0 V^2 = 0.
25 C_0 V^2 \).
\( W = 0.
25 (0.
025 \times 10^{-6}) (100)^2 = 0.
25 (0.
25 \times 10^{-3}) = 0.
0625 \times 10^{-3} \) J \( = 62.
5 \) \( \mu \)J.
Final Answer: The final answer is 62.
5.