Question

A parallel plate capacitor of capacitance \(C\) is connected to a battery and is charged to a potential difference \(V\). Another capacitor of capacitance \(2C\) is similarly charged to a potential difference \(2V\). The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

• A. Zero
• B. \(\dfrac{3}{2}CV^2\)
• C. \(\dfrac{25}{6}CV^2\)
• D. \(\dfrac{9}{2}CV^2\)

Hint:

For capacitors joined after charging: \[ U_f=\frac{Q_{\text{net}}^2}{2C_{\text{eq}}} \] Account carefully for charge signs.

Answer 1

The Correct Option is B

Step 1: Initial charges: \[ Q_1 = CV,\qquad Q_2 = 2C(2V)=4CV \]
Step 2: Since connected oppositely, net charge: \[ Q_{\text{net}} = 4CV - CV = 3CV \]
Step 3: Equivalent capacitance: \[ C_{\text{eq}} = C + 2C = 3C \]
Step 4: Final voltage: \[ V_f = \frac{Q_{\text{net}}}{C_{\text{eq}}} = V \]
Step 5: Final energy: \[ U_f = \frac{1}{2}C_{\text{eq}}V_f^2 = \frac{1}{2}(3C)V^2 = \frac{3}{2}CV^2 \]