Condenser `A` has a capacity of `15 muF` when it is filled with a medium of dielectric constant `15`. Another condenser `B` has a capacity `1muF` with air between the plates. Both are charged separately by a battery of `100V`. After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is

Options

A. 400 V
B. 800 V
C. 1200 V
D. 1600 V

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Calculate the charge on each capacitor before they are connected.

1. **Capacitor A** has a capacitance \( C_A = 15 \mu F \) and is charged to a potential \( V = 100 V \).
- The charge \( Q_A \) on capacitor A can be calculated using the formula:
\[ Q_A = C_A \times V = 15 \mu F \times 100 V = 1500 \mu C \]

2. **Capacitor B** has a capacitance \( C_B = 1 \mu F \) and is also charged to \( V = 100 V \).
- The charge \( Q_B \) on capacitor B can be calculated as:
\[ Q_B = C_B \times V = 1 \mu F \times 100 V = 100 \mu C \]

### Step 2: Calculate the total charge when both capacitors are connected in parallel.

- The total charge \( Q_{total} \) when the two capacitors are connected in parallel is the sum of the charges on both capacitors:
\[ Q_{total} = Q_A + Q_B = 1500 \mu C + 100 \mu C = 1600 \mu C \]

### Step 3: Determine the capacitance of capacitor A after the dielectric is removed.

- When the dielectric is removed, the capacitance of capacitor A changes. The capacitance with the dielectric is \( C_A = 15 \mu F \) and the dielectric constant \( k = 15 \). The capacitance without the dielectric is:
\[ C_{A, \text{air}} = \frac{C_A}{k} = \frac{15 \mu F}{15} = 1 \mu F \]

### Step 4: Calculate the equivalent capacitance when both capacitors are connected in parallel.

- The capacitance of capacitor B remains the same since it was already in air:
\[ C_B = 1 \mu F \]

- The equivalent capacitance \( C_{eq} \) when both capacitors are connected in parallel is:
\[ C_{eq} = C_{A, \text{air}} + C_B = 1 \mu F + 1 \mu F = 2 \mu F \]

### Step 5: Calculate the common potential across the capacitors after they are connected.

- The common potential \( V_{common} \) across the capacitors can be found using the total charge and the equivalent capacitance:
\[ V_{common} = \frac{Q_{total}}{C_{eq}} = \frac{1600 \mu C}{2 \mu F} = 800 V \]

### Final Answer:
The common potential when both capacitors are connected in parallel is \( 800 V \).

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