Two parallel plate condenser A and B having capacitances of `1 mu F and 5 mu F` are charged separately to the same potential of 100V. Now the positive plate of A is connected to the negative plate of B and the negative plate of A to the positive plate of B.
Common potential is

Options

A. `200/5V`
B. `200/3 V`
C. `200/4 V`
D. `200/7 V`

Answer

To solve the problem of finding the common potential after connecting two capacitors in the described manner, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Capacitance of capacitor A, \( C_A = 1 \mu F = 1 \times 10^{-6} F \)
- Capacitance of capacitor B, \( C_B = 5 \mu F = 5 \times 10^{-6} F \)
- Initial potential across both capacitors, \( V_0 = 100 V \)

2. **Calculate Initial Charges:**
- The charge stored in capacitor A, \( Q_A = C_A \times V_0 = 1 \mu F \times 100 V = 1 \times 10^{-6} F \times 100 V = 1 \times 10^{-4} C \)
- The charge stored in capacitor B, \( Q_B = C_B \times V_0 = 5 \mu F \times 100 V = 5 \times 10^{-6} F \times 100 V = 5 \times 10^{-4} C \)

3. **Determine the Initial Charge Before Connection:**
- Since capacitor A is positively charged and capacitor B is negatively charged when connected, the total initial charge \( Q_{initial} \) is given by:
\[ Q_{initial} = Q_A - Q_B = 1 \times 10^{-4} C - 5 \times 10^{-4} C = -4 \times 10^{-4} C \]

4. **Calculate the Total Capacitance After Connection:**
- When capacitors are connected in this manner, they act in parallel. Therefore, the total capacitance \( C_{eq} \) is:
\[ C_{eq} = C_A + C_B = 1 \mu F + 5 \mu F = 6 \mu F = 6 \times 10^{-6} F \]

5. **Apply Conservation of Charge:**
- The final charge \( Q_{final} \) after redistribution will be equal to the initial charge:
\[ Q_{final} = C_{eq} \times V' = -4 \times 10^{-4} C \]
- Rearranging gives:
\[ V' = \frac{Q_{final}}{C_{eq}} = \frac{-4 \times 10^{-4} C}{6 \times 10^{-6} F} = -\frac{4}{6} \times 10^{2} V = -\frac{2}{3} \times 100 V \]

6. **Calculate the Common Potential:**
- Thus, the common potential \( V' \) is:
\[ V' = -\frac{200}{3} V \approx -66.67 V \]

### Final Answer:
The common potential after connecting the capacitors is \( -\frac{200}{3} V \) or approximately \( -66.67 V \).