A capacitor of capacity `C` is connected with a battery of potential `V` in parallel. The distance between its plates is reduced to half at one, assuming that the charge remains the same. Then to charge the capacitance upto the potential `V` again, the energy given by the battery will be

Options

A. `CV^(2)//4`
B. `CV^(2)//2`
C. `2CV^(2)//4`
D. `CV^(2)`

Answer

To solve the problem step by step, we will analyze the situation involving the capacitor, its capacitance, charge, and the energy supplied by the battery.

### Step 1: Understand the Initial Conditions
- A capacitor of capacitance \( C \) is connected to a battery of potential \( V \).
- The initial charge \( Q \) on the capacitor can be calculated using the formula:
\[ Q = C \cdot V \]

### Step 2: Change in Capacitance
- When the distance between the plates of the capacitor is reduced to half, the new capacitance \( C' \) can be calculated using the formula:
\[ C' = \frac{A \epsilon_0}{D/2} = 2C \]
- Thus, the new capacitance \( C' \) is double the original capacitance.

### Step 3: Charge on the New Capacitor
- Since the capacitor is still connected to the same battery, the charge on the new capacitor \( Q' \) when fully charged will be:
\[ Q' = C' \cdot V = 2C \cdot V = 2Q \]
- Here, \( Q' \) is twice the initial charge \( Q \).

### Step 4: Charge Adjustment
- To maintain the same charge \( Q \) as before, we need to remove some charge from the capacitor. The amount of charge that needs to be removed \( \Delta Q \) is:
\[ \Delta Q = Q' - Q = 2Q - Q = Q \]

### Step 5: Energy Supplied by the Battery
- The energy \( W \) supplied by the battery to remove this charge \( \Delta Q \) at potential \( V \) is given by:
\[ W = \Delta Q \cdot V = Q \cdot V \]
- Substituting \( Q = C \cdot V \):
\[ W = (C \cdot V) \cdot V = C V^2 \]

### Step 6: Final Result
- Therefore, the energy given by the battery to charge the capacitor up to potential \( V \) again is:
\[ W = C V^2 \]