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 V
B. 200 V
C. 600 V
D. 400 V

Answer

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

### Step 1: Understand the Configuration
We have two capacitors:
- Capacitor 1 (C1) = 2 µF
- Capacitor 2 (C2) = 3 µF

They are connected in series, with the outer plate of C1 at 1000 V and the outer plate of C2 grounded (0 V).

### Step 2: Use the Concept of Series Capacitors
In a series connection, the charge (Q) on each capacitor is the same. The potential difference across each capacitor can be expressed as:
- \( V_1 = \frac{Q}{C_1} \)
- \( V_2 = \frac{Q}{C_2} \)

The total voltage across the series combination is the sum of the voltages across each capacitor:
\[ V_{total} = V_1 + V_2 \]

### Step 3: Write the Voltage Equations
Given that the outer plate of C1 is at 1000 V and the outer plate of C2 is at 0 V, we can express the total voltage as:
\[ 1000 V - V_1 = V_2 - 0 V \]
Thus, we can write:
\[ V_1 + V_2 = 1000 \]

### Step 4: Relate Voltages to Charges
From the equations of voltage across the capacitors, we can substitute \( V_1 \) and \( V_2 \):
\[ V_1 = \frac{Q}{C_1} = \frac{Q}{2 \times 10^{-6}} \]
\[ V_2 = \frac{Q}{C_2} = \frac{Q}{3 \times 10^{-6}} \]

### Step 5: Substitute and Solve
Substituting these into the total voltage equation:
\[ \frac{Q}{2 \times 10^{-6}} + \frac{Q}{3 \times 10^{-6}} = 1000 \]

To solve for Q, we first find a common denominator:
\[ \frac{3Q + 2Q}{6 \times 10^{-6}} = 1000 \]
\[ \frac{5Q}{6 \times 10^{-6}} = 1000 \]

Now, multiply both sides by \( 6 \times 10^{-6} \):
\[ 5Q = 1000 \times 6 \times 10^{-6} \]
\[ 5Q = 6 \times 10^{-3} \]
\[ Q = \frac{6 \times 10^{-3}}{5} = 1.2 \times 10^{-3} \, C \]

### Step 6: Calculate the Voltages
Now, we can find \( V_1 \) and \( V_2 \):
\[ V_1 = \frac{Q}{C_1} = \frac{1.2 \times 10^{-3}}{2 \times 10^{-6}} = 600 \, V \]
\[ V_2 = \frac{Q}{C_2} = \frac{1.2 \times 10^{-3}}{3 \times 10^{-6}} = 400 \, V \]

### Step 7: Find the Potential on Inner Plates
The potential on the inner plate of C1 (V1) is 600 V, and the potential on the inner plate of C2 (V2) is 400 V.

### Final Answer
- The potential on the inner plate of the first capacitor (C1) is **600 V**.
- The potential on the inner plate of the second capacitor (C2) is **400 V**.