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. (a)5
B. (b)10
C. (c)25
D. (d)50

Explanation

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

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

Given:
- \( C = 5 \mu F = 5 \times 10^{-6} F \)
- \( V = 100 V \)

Calculating the charge:
\[ Q = 5 \times 10^{-6} F \times 100 V = 5 \times 10^{-4} C = 500 \mu C \]

### Step 2: Understand the effect of doubling the distance between the plates
When the distance between the plates of a capacitor is doubled, the capacitance \( C \) changes. The capacitance \( C \) is inversely proportional to the distance \( d \) between the plates:
\[ C' = \frac{C}{2} \]
So the new capacitance \( C' \) when the distance is doubled becomes:
\[ C' = \frac{5 \mu F}{2} = 2.5 \mu F \]

### Step 3: Calculate the initial potential energy of the capacitor
The potential energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]

Calculating the initial potential energy \( U_i \):
\[ U_i = \frac{1}{2} \times 5 \times 10^{-6} F \times (100 V)^2 \]
\[ U_i = \frac{1}{2} \times 5 \times 10^{-6} \times 10000 = \frac{1}{2} \times 5 \times 10^{-2} = 2.5 \times 10^{-2} J = 25 mJ \]

### Step 4: Calculate the final potential energy after the distance is doubled
Now, we calculate the potential energy \( U_f \) with the new capacitance \( C' \):
\[ U_f = \frac{1}{2} C' V'^2 \]
Since the charge \( Q \) remains constant, we can find the new voltage \( V' \) using:
\[ Q = C' V' \implies V' = \frac{Q}{C'} = \frac{500 \times 10^{-6}}{2.5 \times 10^{-6}} = 200 V \]

Now calculating the final potential energy:
\[ U_f = \frac{1}{2} \times 2.5 \times 10^{-6} F \times (200 V)^2 \]
\[ U_f = \frac{1}{2} \times 2.5 \times 10^{-6} \times 40000 = \frac{1}{2} \times 0.1 = 0.05 J = 50 mJ \]

### Step 5: Calculate the work done to separate the plates
The work done \( W \) is the difference in potential energy:
\[ W = U_f - U_i \]
\[ W = 50 mJ - 25 mJ = 25 mJ \]

### Final Answer:
The minimum work needed to pull the plates of the capacitor apart so that the distance between them doubles is \( \boxed{25 \, mJ} \).

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