A capacitor of capacitance `5muF` is charged to a potential difference 100V and then diconnected from the powerr supply. The minimum work needed to pull the plates of the capacitor apart so that the distance between them doubles is (in mJ).

Options

A. 5
B. 10
C. 25
D. 50

Explanation

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

### Step 1: Calculate the charge on the capacitor
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \cdot V \]
Where:
- \( C = 5 \mu F = 5 \times 10^{-6} F \)
- \( V = 100 V \)

Substituting the values:
\[ Q = 5 \times 10^{-6} F \times 100 V = 5 \times 10^{-4} C \]

### Step 2: Understand the change in distance between the plates
Initially, let the distance between the plates be \( D \). After pulling the plates apart, the new distance becomes \( 2D \).

### Step 3: Calculate the initial and final capacitance
The capacitance \( C \) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{D} \]
Where:
- \( \varepsilon_0 \) is the permittivity of free space
- \( A \) is the area of the plates

Initially, the capacitance \( C_1 \) is:
\[ C_1 = 5 \mu F = 5 \times 10^{-6} F \]

When the distance is doubled, the new capacitance \( C_2 \) is:
\[ C_2 = \frac{\varepsilon_0 A}{2D} = \frac{C_1}{2} = \frac{5 \mu F}{2} = 2.5 \mu F = 2.5 \times 10^{-6} F \]

### Step 4: Calculate the initial and final energy stored in the capacitor
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]

**Initial energy \( U_1 \)**:
\[ U_1 = \frac{1}{2} C_1 V^2 = \frac{1}{2} \cdot 5 \times 10^{-6} F \cdot (100 V)^2 \]
\[ U_1 = \frac{1}{2} \cdot 5 \times 10^{-6} \cdot 10000 = \frac{5 \times 10^{-6} \cdot 10000}{2} = 25 \times 10^{-3} J = 25 mJ \]

**Final energy \( U_2 \)**:
\[ U_2 = \frac{1}{2} C_2 V_2^2 \]
To find \( V_2 \), we use the relation \( Q = C_2 V_2 \):
\[ V_2 = \frac{Q}{C_2} = \frac{5 \times 10^{-4}}{2.5 \times 10^{-6}} = 200 V \]
Now calculate \( U_2 \):
\[ U_2 = \frac{1}{2} \cdot 2.5 \times 10^{-6} F \cdot (200 V)^2 \]
\[ U_2 = \frac{1}{2} \cdot 2.5 \times 10^{-6} \cdot 40000 = 50 \times 10^{-3} J = 50 mJ \]

### Step 5: Calculate the work done
The work done \( W \) to pull the plates apart is the difference in energy:
\[ W = U_2 - U_1 = 50 mJ - 25 mJ = 25 mJ \]

### Final Answer
The minimum work needed to pull the plates of the capacitor apart so that the distance between them doubles is **25 mJ**.

---