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 ismilarly 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 poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Options

A. zero
B. `(25 CV^(2))/(6)`
C. `(3 CV^(2))/(2)`
D. `(9 CV^(2))/(2)`

Answer

To find the final energy of the configuration when two capacitors are connected in parallel with opposite polarity, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the Initial Charges:**
- For the first capacitor (capacitance \(C\) and voltage \(V\)):
\[ Q_1 = C \cdot V \]
- For the second capacitor (capacitance \(2C\) and voltage \(2V\)):
\[ Q_2 = 2C \cdot 2V = 4CV \]

2. **Determine the Net Charge:**
- When the two capacitors are connected in parallel with opposite polarity, the net charge \(Q_{\text{net}}\) is given by:
\[ Q_{\text{net}} = Q_1 - Q_2 = CV - 4CV = -3CV \]
- The negative sign indicates that the net charge is in the direction of the larger charge.

3. **Calculate the Equivalent Capacitance:**
- The equivalent capacitance \(C_{\text{eq}}\) when capacitors are connected in parallel is the sum of their capacitances:
\[ C_{\text{eq}} = C + 2C = 3C \]

4. **Find the Final Voltage:**
- The final voltage \(V_f\) across the equivalent capacitance can be calculated using the formula:
\[ V_f = \frac{Q_{\text{net}}}{C_{\text{eq}}} = \frac{-3CV}{3C} = -V \]
- The negative sign indicates that the voltage is in the opposite direction to the initial voltage.

5. **Calculate the Final Energy:**
- The energy \(U\) stored in the capacitors can be calculated using the formula:
\[ U = \frac{Q_{\text{net}}^2}{2C_{\text{eq}}} \]
- Substituting the values we have:
\[ U = \frac{(-3CV)^2}{2 \cdot 3C} = \frac{9C^2V^2}{6C} = \frac{3}{2} CV^2 \]

### Final Answer:
The final energy of the configuration is:
\[ \frac{3}{2} CV^2 \]