The plates of a parallel plate capacitor of capacity `50 mu C` are charged to a potential of `100` volts and then separated from each other so that the distance between then is doubled. How much is the energy spent in doing so

Options

A. `25 xx 10^(-2) J`
B. `-12.5 xx 10^(-2) J`
C. `-25 xx 10^(-2) J`
D. `12.5 xx 10^(-2) J`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We have a parallel plate capacitor with:
- Initial capacitance \( C = 50 \, \mu F \) (microfarads)
- Voltage \( V = 100 \, V \) (volts)

### Step 2: Calculate the initial charge on the capacitor
The charge \( Q \) on the capacitor is given by the formula:
\[ Q = C \cdot V \]
Substituting the values:
\[ Q = 50 \times 10^{-6} \, F \times 100 \, V = 5 \times 10^{-3} \, C \]

### Step 3: Calculate the initial energy stored in the capacitor
The energy \( U_1 \) stored in a capacitor is given by:
\[ U = \frac{1}{2} \cdot C \cdot V^2 \]
Substituting the values:
\[ U_1 = \frac{1}{2} \cdot 50 \times 10^{-6} \, F \cdot (100 \, V)^2 \]
Calculating:
\[ U_1 = \frac{1}{2} \cdot 50 \times 10^{-6} \cdot 10000 = \frac{1}{2} \cdot 50 \times 10^{-2} = 25 \times 10^{-3} \, J = 0.025 \, J \]

### Step 4: Determine the new capacitance after doubling the distance
When the distance between the plates is doubled, the new capacitance \( C_2 \) becomes:
\[ C_2 = \frac{C}{2} = \frac{50 \, \mu F}{2} = 25 \, \mu F \]

### Step 5: Calculate the final energy stored in the capacitor
Since the battery is disconnected, the charge remains the same. The new energy \( U_2 \) stored in the capacitor is given by:
\[ U_2 = \frac{Q^2}{2C_2} \]
Substituting the values:
\[ U_2 = \frac{(5 \times 10^{-3})^2}{2 \cdot 25 \times 10^{-6}} = \frac{25 \times 10^{-6}}{50 \times 10^{-6}} = \frac{25}{50} = 0.5 \, J \]

### Step 6: Calculate the energy spent in separating the plates
The energy spent in doing so is the difference in energy before and after:
\[ \text{Energy spent} = U_1 - U_2 \]
Substituting the values:
\[ \text{Energy spent} = 0.025 \, J - 0.5 \, J = -0.475 \, J \]

### Final Answer
The energy spent in separating the plates is \( 0.025 \, J - 0.0125 \, J = 0.0125 \, J \).

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