A parallel plate capacitor of capacity 100`mu F` is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is

Options

A. A) `62.5 xx 10^(-3)`
B. B) `12.5 xx 10^(-3)`
C. C) `1.25 xx 10^(-3)`
D. D) `0.125 xx 10^(-3)`

Answer

To solve the problem step by step, we will calculate the initial and final energy of the capacitor and then find the additional energy given by the battery when the distance between the plates is doubled.

### Step 1: Calculate the Initial Energy of the Capacitor
The initial energy \( U_1 \) stored in a capacitor is given by the formula:
\[ U_1 = \frac{1}{2} C V^2 \]
Where:
- \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- \( V = 50 \, V \)

Substituting the values:
\[ U_1 = \frac{1}{2} \times (100 \times 10^{-6}) \times (50)^2 \]
Calculating \( (50)^2 \):
\[ (50)^2 = 2500 \]
Now substituting this back:
\[ U_1 = \frac{1}{2} \times (100 \times 10^{-6}) \times 2500 \]
\[ U_1 = \frac{1}{2} \times 250 \times 10^{-3} = 125 \times 10^{-3} \, J = 0.125 \, J \]

### Step 2: Determine the New Capacitance After Doubling the Distance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{d} \]
When the distance \( d \) is doubled, the new capacitance \( C' \) becomes:
\[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} C \]
Thus:
\[ C' = \frac{1}{2} \times 100 \, \mu F = 50 \, \mu F = 50 \times 10^{-6} \, F \]

### Step 3: Calculate the Final Energy of the Capacitor
The final energy \( U_2 \) stored in the capacitor when the battery is still connected is:
\[ U_2 = \frac{1}{2} C' V^2 \]
Substituting the values:
\[ U_2 = \frac{1}{2} \times (50 \times 10^{-6}) \times (50)^2 \]
Calculating:
\[ U_2 = \frac{1}{2} \times (50 \times 10^{-6}) \times 2500 \]
\[ U_2 = \frac{1}{2} \times 125 \times 10^{-3} = 62.5 \times 10^{-3} \, J = 0.0625 \, J \]

### Step 4: Calculate the Additional Energy Given by the Battery
The additional energy given by the battery is the difference between the initial and final energy:
\[ \text{Additional Energy} = U_1 - U_2 \]
Substituting the values:
\[ \text{Additional Energy} = (0.125 \, J) - (0.0625 \, J) = 0.0625 \, J \]

### Final Answer
The additional energy given by the battery to the capacitor is:
\[ \text{Additional Energy} = 0.0625 \, J \]