Two capacitors of capacitance `2 muF` and `3 muF` are joined in series. Outer plate first capacitor is at `1000` volt and outer plate of second capacitor is earthed (grounded). Now the potential on inner plate of each capacitor will be

Options

A. `700` Volt
B. `200` Volt
C. `600` Volt
D. `400` Volt

Explanation

To solve the problem of finding the potential on the inner plates of two capacitors connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Capacitors and Their Capacitances**:
- We have two capacitors:
- \( C_1 = 2 \, \mu F \)
- \( C_2 = 3 \, \mu F \)

2. **Understand the Configuration**:
- The outer plate of the first capacitor is at \( 1000 \, V \).
- The outer plate of the second capacitor is grounded (at \( 0 \, V \)).
- The capacitors are connected in series.

3. **Determine the Voltage Across Each Capacitor**:
- Let \( V \) be the potential on the inner plate of the first capacitor (between \( C_1 \) and \( C_2 \)).
- The voltage across the first capacitor \( C_1 \) is given by:
\[ V_{C1} = 1000 \, V - V \]
- The voltage across the second capacitor \( C_2 \) is:
\[ V_{C2} = V - 0 \, V = V \]

4. **Use the Relationship of Charge in Series**:
- In series, the charge \( Q \) on both capacitors is the same:
\[ Q = C_1 \cdot V_{C1} = C_2 \cdot V_{C2} \]

5. **Substitute the Values**:
- For \( C_1 \):
\[ Q = 2 \times 10^{-6} \cdot (1000 - V) \]
- For \( C_2 \):
\[ Q = 3 \times 10^{-6} \cdot V \]

6. **Set the Equations Equal**:
\[ 2 \times 10^{-6} (1000 - V) = 3 \times 10^{-6} V \]

7. **Simplify the Equation**:
- Cancel \( 10^{-6} \) from both sides:
\[ 2(1000 - V) = 3V \]
- Expanding gives:
\[ 2000 - 2V = 3V \]

8. **Combine Like Terms**:
\[ 2000 = 5V \]

9. **Solve for \( V \)**:
\[ V = \frac{2000}{5} = 400 \, V \]

10. **Conclusion**:
- The potential on the inner plate of the first capacitor is \( 400 \, V \).
- The potential on the inner plate of the second capacitor (which is the same as the inner plate of the first capacitor) is also \( 400 \, V \).

### Final Answer:
- The potential on the inner plate of each capacitor is \( 400 \, V \).