A parallel plate capacitor of capacity `C_(0)` is charged to a potential `V_(0), E_(1)` is the energy stored in the capacitor when the battery is disconnected and the plate separation is doubled, and `E_(2)` is the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled. find the ratio `E_(1)//E_(2)`.

Explanation

To solve the problem, we need to find the ratio \( \frac{E_1}{E_2} \), where \( E_1 \) is the energy stored in the capacitor when the battery is disconnected and the plate separation is doubled, and \( E_2 \) is the energy stored when the battery remains connected while the plate separation is doubled.

### Step-by-Step Solution:

1. **Initial Energy Calculation**:
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
When the capacitor is charged to a potential \( V_0 \) with capacitance \( C_0 \), the initial energy \( E_0 \) is:
\[ E_0 = \frac{1}{2} C_0 V_0^2 \]

2. **Energy When Battery is Disconnected (Finding \( E_1 \))**:
After disconnecting the battery, the charge \( Q \) on the capacitor remains constant. The charge can be expressed as:
\[ Q = C_0 V_0 \]
When the plate separation is doubled, the new capacitance \( C' \) becomes:
\[ C' = \frac{\epsilon_0 A}{2D} = \frac{C_0}{2} \]
The energy stored in the capacitor after doubling the separation is:
\[ E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2 \cdot \frac{C_0}{2}} = \frac{(C_0 V_0)^2}{C_0} = \frac{C_0 V_0^2}{2} \cdot 2 = C_0 V_0^2 \]

3. **Energy When Battery is Connected (Finding \( E_2 \))**:
When the battery remains connected, the potential \( V \) across the capacitor remains \( V_0 \). The new capacitance after doubling the separation is still \( C' = \frac{C_0}{2} \). The energy stored in this case is:
\[ E_2 = \frac{1}{2} C' V_0^2 = \frac{1}{2} \cdot \frac{C_0}{2} \cdot V_0^2 = \frac{C_0 V_0^2}{4} \]

4. **Finding the Ratio \( \frac{E_1}{E_2} \)**:
Now we can find the ratio of the energies:
\[ \frac{E_1}{E_2} = \frac{C_0 V_0^2}{\frac{C_0 V_0^2}{4}} = \frac{C_0 V_0^2 \cdot 4}{C_0 V_0^2} = 4 \]

### Final Result:
The ratio \( \frac{E_1}{E_2} \) is:
\[ \frac{E_1}{E_2} = 4 \]