A `10 muF` capacitor is charged to a potential difference of `50 V` and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes `20` volt. The capacitance of second capacitor is

Options

A. `15 mu `F
B. `20 muF`
C. `10 mu` F
D. `30 mu ` F

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial charge on the first capacitor
The charge \( Q_1 \) on the first capacitor can be calculated using the formula:
\[ Q_1 = C_1 \times V_1 \]
where \( C_1 = 10 \, \mu F \) and \( V_1 = 50 \, V \).

\[ Q_1 = 10 \times 10^{-6} \, F \times 50 \, V = 500 \, \mu C \]

### Step 2: Determine the final charge on the first capacitor
After connecting the charged capacitor to the uncharged capacitor in parallel, the common potential difference across both capacitors becomes \( V_f = 20 \, V \). The charge \( Q_1' \) on the first capacitor after connection is given by:
\[ Q_1' = C_1 \times V_f \]
Substituting the values:
\[ Q_1' = 10 \times 10^{-6} \, F \times 20 \, V = 200 \, \mu C \]

### Step 3: Calculate the charge on the second capacitor
Since charge is conserved, the total charge before and after connecting the capacitors must be equal. Therefore, the charge \( Q_2 \) on the second capacitor can be calculated as:
\[ Q = Q_1 = Q_1' + Q_2 \]
Rearranging gives:
\[ Q_2 = Q - Q_1' = 500 \, \mu C - 200 \, \mu C = 300 \, \mu C \]

### Step 4: Calculate the capacitance of the second capacitor
The charge on the second capacitor is given by:
\[ Q_2 = C_2 \times V_f \]
Rearranging for \( C_2 \):
\[ C_2 = \frac{Q_2}{V_f} \]
Substituting the values:
\[ C_2 = \frac{300 \, \mu C}{20 \, V} = \frac{300 \times 10^{-6} \, C}{20 \, V} = 15 \, \mu F \]

### Final Answer
The capacitance of the second capacitor is \( 15 \, \mu F \).

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