The plates of a parallel plate capacitor have surface area .A. and are initially separated by a distance .d.. They are connected to a battery of voltage `v_0` Now, the plates of the capacitor are pulled apart w a separation 2d. Then, increase in the energy of the battery is

Options

A. `(in_(0)AV_0^2)/(4d)`
B. `(in_(0)AV_0^2)/(2d)`
C. `(in_(0)AV_0^2)/(8d)`
D. there will be no change in the energy of the battery

Answer

To solve the problem step by step, we will analyze the situation of the parallel plate capacitor before and after the plates are pulled apart.

### Step 1: Understand the initial conditions
- The initial capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\epsilon_0 A}{d} \]
where \( \epsilon_0 \) is the permittivity of free space, \( A \) is the surface area of the plates, and \( d \) is the distance between the plates.

### Step 2: Determine the initial energy stored in the capacitor
- The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} Q V \]
Since the capacitor is connected to a battery of voltage \( V_0 \), the charge \( Q \) on the capacitor is:
\[ Q = C V_0 = \frac{\epsilon_0 A}{d} V_0 \]
- Thus, the initial energy \( U_i \) is:
\[ U_i = \frac{1}{2} Q V_0 = \frac{1}{2} \left(\frac{\epsilon_0 A}{d} V_0\right) V_0 = \frac{\epsilon_0 A V_0^2}{2d} \]

### Step 3: Analyze the new conditions after pulling the plates apart
- When the plates are pulled apart to a distance of \( 2d \), the new capacitance \( C' \) becomes:
\[ C' = \frac{\epsilon_0 A}{2d} \]

### Step 4: Calculate the new energy stored in the capacitor
- The charge \( Q \) remains the same since the battery is still connected, so:
\[ U_f = \frac{1}{2} \frac{Q^2}{C'} = \frac{1}{2} \frac{Q^2}{\frac{\epsilon_0 A}{2d}} = \frac{Q^2 \cdot 2d}{2 \epsilon_0 A} \]
- Substituting \( Q = \frac{\epsilon_0 A}{d} V_0 \):
\[ U_f = \frac{1}{2} \frac{\left(\frac{\epsilon_0 A}{d} V_0\right)^2}{\frac{\epsilon_0 A}{2d}} = \frac{1}{2} \cdot \frac{\epsilon_0^2 A^2 V_0^2}{d^2} \cdot \frac{2d}{\epsilon_0 A} = \frac{\epsilon_0 A V_0^2}{d} \]

### Step 5: Calculate the increase in energy of the battery
- The increase in energy \( \Delta U \) is given by:
\[ \Delta U = U_f - U_i \]
- Substituting the expressions for \( U_f \) and \( U_i \):
\[ \Delta U = \frac{\epsilon_0 A V_0^2}{d} - \frac{\epsilon_0 A V_0^2}{2d} = \frac{\epsilon_0 A V_0^2}{d} - \frac{\epsilon_0 A V_0^2}{2d} = \frac{\epsilon_0 A V_0^2}{2d} \]

### Final Answer:
The increase in the energy of the battery is:
\[ \Delta U = \frac{\epsilon_0 A V_0^2}{2d} \]