A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B as shown in Figure (a). Then it is

Question

A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery B as shown in Figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance C = 900 pF as shown in Figure (b). The electrostatic energy stored by the system (b) is:

Accepted Answer

1. **Initial State (Figure a):**
   - Capacitance $C_1 = 900 	ext{ pF} = 9 	imes 10^{-10} 	ext{ F}$.
   - Voltage $V_1 = 100 	ext{ V}$.
   - Initial Energy stored: $U_i = \frac{1}{2}C_1 V_1^2 = \frac{1}{2}(9 	imes 10^{-10})(100)^2 = 4.5 	imes 10^{-6} 	ext{ J}$.
   - Charge stored: $Q = C_1 V_1 = (900 	imes 10^{-12})(100) = 9 	imes 10^{-8} 	ext{ C}$.

2. **Final State (Figure b):**
   - The charged capacitor $C_1$ is disconnected and connected in parallel to an identical uncharged capacitor $C_2 = 900 	ext{ pF}$.
   - **Charge Redistribution:** The total charge $Q$ remains conserved but distributes equally between the two identical capacitors. Thus, charge on each is $Q' = Q/2$.
   - **System Capacitance:** The two capacitors are in parallel, so equivalent capacitance $C_{eq} = C_1 + C_2 = 2C = 1800 	ext{ pF}$.
   - **Final Energy:** The energy stored in the system is $U_f = \frac{Q^2}{2C_{eq}}$.
   - Substituting $C_{eq} = 2C_1$: $U_f = \frac{Q^2}{2(2C_1)} = \frac{1}{2} \left( \frac{Q^2}{2C_1} ight) = \frac{1}{2} U_i$.

3. **Calculation:**
   - $U_f = \frac{1}{2} 	imes (4.5 	imes 10^{-6} 	ext{ J}) = 2.25 	imes 10^{-6} 	ext{ J}$.

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