Question

A parallel plate capacitor of capacitance 10 \(\mu\)F is charged by a 220 V supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance 12 \(\mu\)F. The loss of electrostatic energy in this process is

• A. 132 mJ
• B. 220 mJ
• C. 66 mJ
• D. 110 mJ

Hint:

A useful shortcut formula for the energy loss when connecting a charged capacitor \(C_1\) (at potential \(V_1\)) to an uncharged capacitor \(C_2\) is: \( \Delta U_{loss} = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} V_1^2 \). This formula directly calculates the energy dissipated as heat during the charge redistribution.

Answer 1

The Correct Option is A

Step 1: Calculate the initial energy stored in the first capacitor.
Let \(C_1 = 10 \mu\text{F} = 10 \times 10^{-6}\) F and \(V_1 = 220\) V.
The initial energy stored in \(C_1\) is \( U_{initial} = \frac{1}{2}C_1 V_1^2 \).
\( U_{initial} = \frac{1}{2} (10 \times 10^{-6}) (220)^2 = 5 \times 10^{-6} \times 48400 = 0.242 \) J.
The initial charge on \(C_1\) is \( Q_1 = C_1 V_1 = (10 \times 10^{-6}) \times 220 = 2200 \times 10^{-6} = 2.2 \times 10^{-3} \) C.
Step 2: Calculate the state after connecting the capacitors.
When \(C_1\) is connected to an uncharged capacitor \(C_2 = 12 \mu\text{F}\), the total charge Q is conserved and redistributes.
The total capacitance of the parallel combination is \( C_{total} = C_1 + C_2 = 10 + 12 = 22 \mu\text{F} \).
The total charge is \( Q_{total} = Q_1 = 2.2 \times 10^{-3} \) C.
The final common potential (V_final) across the combination is \( V_{final} = \frac{Q_{total}}{C_{total}} \).
\( V_{final} = \frac{2.2 \times 10^{-3}}{22 \times 10^{-6}} = \frac{2.2}{22} \times 10^3 = 0.1 \times 10^3 = 100 \) V.
Step 3: Calculate the final energy stored in the system.
The final energy is \( U_{final} = \frac{1}{2}C_{total} V_{final}^2 \).
\( U_{final} = \frac{1}{2} (22 \times 10^{-6}) (100)^2 = 11 \times 10^{-6} \times 10000 = 0.11 \) J.
Step 4: Calculate the loss of energy.
Energy Loss \( \Delta U = U_{initial} - U_{final} \).
\( \Delta U = 0.242 \text{ J} - 0.11 \text{ J} = 0.132 \) J.
To express this in millijoules (mJ), we multiply by 1000.
\( \Delta U = 0.132 \times 1000 = 132 \) mJ.