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 up to the potential `V` again, the energy given by the battery will be

Options

A. `CV^(2)//4`
B. `CV^(2)//2`
C. `3CV^(2)//4`
D. `CV^(2)`

Explanation

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
Initially, we have a capacitor with capacitance \( C \) connected to a battery with potential \( V \). The charge \( Q \) on the capacitor can be expressed as:
\[ Q = C \cdot V \]

### Step 2: Determine the Effect of Reducing the Distance
When the distance between the plates of the capacitor is reduced to half, the new distance \( D' \) becomes:
\[ D' = \frac{D}{2} \]
The capacitance of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 \cdot A}{D} \]
where \( A \) is the area of the plates and \( \varepsilon_0 \) is the permittivity of free space. When the distance is halved, the new capacitance \( C' \) becomes:
\[ C' = \frac{\varepsilon_0 \cdot A}{D'} = \frac{\varepsilon_0 \cdot A}{D/2} = 2 \cdot \frac{\varepsilon_0 \cdot A}{D} = 2C \]

### Step 3: Calculate the New Charge on the Capacitor
Since the charge remains the same during the process, we have:
\[ Q' = Q = C \cdot V \]
However, when the capacitance changes to \( C' \), the new charge \( Q' \) can also be expressed as:
\[ Q' = C' \cdot V = 2C \cdot V \]

### Step 4: Find the Extra Charge Required
The extra charge \( \Delta Q \) that needs to be supplied by the battery to maintain the potential \( V \) is:
\[ \Delta Q = Q' - Q = (2C \cdot V) - (C \cdot V) = C \cdot V \]

### Step 5: Calculate the Work Done by the Battery
The work done \( W \) by the battery to supply this extra charge \( \Delta Q \) at potential \( V \) is given by:
\[ W = \Delta Q \cdot V = (C \cdot V) \cdot V = C \cdot V^2 \]

### Conclusion
Thus, the energy given by the battery to charge the capacitor back to the potential \( V \) after the distance between the plates has been halved is:
\[ \boxed{C \cdot V^2} \]