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. `10 muF`
B. `20 muF`
C. `30 muF`
D. `15 muF`

Explanation

To solve the problem step by step, we will follow these steps:

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

For the given capacitor:
- \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \)
- \( V = 50 \, V \)

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

### Step 2: Determine the final common potential after connecting the uncharged capacitor
When the charged capacitor is connected to an uncharged capacitor in parallel, the total charge remains the same, but the potential difference changes. The new common potential \( V_f \) is given as:
\[ V_f = 20 \, V \]

### Step 3: Set up the equation for total charge
The total charge before connecting the second capacitor is equal to the total charge after connecting it. Therefore, we can write:
\[ Q_{initial} = Q_{final} \]
Where \( Q_{final} \) is the total charge after connecting the capacitors:
\[ Q_{final} = C_1 \times V_f + C_2 \times V_f \]
Here, \( C_1 = 10 \, \mu F \) and \( C_2 \) is the capacitance of the second capacitor which we need to find.

Substituting the values:
\[ 500 \, \mu C = 10 \, \mu F \times 20 \, V + C_2 \times 20 \, V \]
\[ 500 \, \mu C = 200 \, \mu C + C_2 \times 20 \, V \]

### Step 4: Solve for \( C_2 \)
Rearranging the equation to isolate \( C_2 \):
\[ 500 \, \mu C - 200 \, \mu C = C_2 \times 20 \, V \]
\[ 300 \, \mu C = C_2 \times 20 \, V \]
Now, solving for \( C_2 \):
\[ C_2 = \frac{300 \, \mu C}{20 \, V} = 15 \, \mu F \]

### Final Answer
The capacitance of the second capacitor is:
\[ C_2 = 15 \, \mu F \]

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