Two capacitors of `2 muF` and `3 mu F` are joined in series. The outer plate of second capacitor is earthed. Find out the potential and charge of the inner plate of each capacitor.

Explanation

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

### Step 1: Identify the Capacitors and Their Values
We have two capacitors:
- Capacitor 1 (C1) = 2 µF
- Capacitor 2 (C2) = 3 µF

### Step 2: Determine the Configuration
The capacitors are connected in series, and the outer plate of the second capacitor is earthed (i.e., it is at 0 volts).

### Step 3: Calculate the Equivalent Capacitance
The formula for the equivalent capacitance (C_eq) of capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \]
Thus,
\[ C_{eq} = \frac{6}{5} \, \mu F = 1.2 \, \mu F \]

### Step 4: Calculate the Total Charge (Q)
The total charge (Q) in the series circuit can be calculated using the formula:
\[ Q = C_{eq} \times V \]
Where V is the total potential difference across the series combination. Since the outer plate of the first capacitor is at 1000 volts and the outer plate of the second capacitor is earthed (0 volts), the total potential difference is:
\[ V = 1000 \, V - 0 \, V = 1000 \, V \]
Now substituting the values:
\[ Q = 1.2 \, \mu F \times 1000 \, V = 1200 \, \mu C \]

### Step 5: Calculate the Potential Difference Across Each Capacitor
1. **For Capacitor 1 (C1)**:
The potential difference (V1) across the first capacitor can be calculated using:
\[ V_1 = \frac{Q}{C_1} = \frac{1200 \, \mu C}{2 \, \mu F} = 600 \, V \]

2. **For Capacitor 2 (C2)**:
The potential difference (V2) across the second capacitor can be calculated using:
\[ V_2 = \frac{Q}{C_2} = \frac{1200 \, \mu C}{3 \, \mu F} = 400 \, V \]

### Step 6: Determine the Potentials of the Inner Plates
- The potential of the inner plate of Capacitor 1 (V1) is:
\[ V_{inner1} = 1000 \, V - V_1 = 1000 \, V - 600 \, V = 400 \, V \]
- The potential of the inner plate of Capacitor 2 (V2) is:
\[ V_{inner2} = 0 \, V + V_2 = 0 \, V + 400 \, V = 400 \, V \]

### Final Results
- Charge on both capacitors: \( Q = 1200 \, \mu C \)
- Potential of the inner plate of Capacitor 1: \( 400 \, V \)
- Potential of the inner plate of Capacitor 2: \( 400 \, V \)