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. `30 mu F`
C. `20 mu F`
D. `10 mu F`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial charge on the first capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the potential difference.

Given:
- Capacitance \( C_1 = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- Potential difference \( V_1 = 50 \, V \)

Calculating the charge:
\[ Q = 10 \times 10^{-6} \, F \times 50 \, V = 500 \, \mu C \]

### Step 2: Determine the total capacitance after connecting the second capacitor
When the charged capacitor is connected in parallel with an uncharged capacitor, the total capacitance \( C_{eq} \) is the sum of the capacitances:
\[ C_{eq} = C_1 + C_2 \]
where \( C_2 \) is the capacitance of the second capacitor.

### Step 3: Use the common potential difference to find the total charge
After connecting the second capacitor, the common potential difference becomes \( V_{common} = 20 \, V \).

The total charge in the system remains the same because charge is conserved. Thus, we can write:
\[ Q = C_{eq} \times V_{common} \]

Substituting the values we have:
\[ 500 \, \mu C = (C_1 + C_2) \times 20 \, V \]

### Step 4: Substitute known values and solve for \( C_2 \)
Substituting \( C_1 = 10 \, \mu F \):
\[ 500 \, \mu C = (10 \, \mu F + C_2) \times 20 \, V \]
Rearranging gives:
\[ 500 = (10 + C_2) \times 20 \]

### Step 5: Isolate \( C_2 \)
Dividing both sides by 20:
\[ 25 = 10 + C_2 \]
Now, subtracting 10 from both sides:
\[ C_2 = 25 - 10 = 15 \, \mu F \]

### Conclusion
The capacitance of the second capacitor \( C_2 \) is \( 15 \, \mu F \).

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