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 dounled. find the ratio `E_(1)//E_(2)`.

Explanation

To find the ratio \( \frac{E_1}{E_2} \) for the given problem, we will follow these steps:

### Step 1: Determine the initial energy stored in the capacitor
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
For the initial capacitor with capacitance \( C_0 \) and potential \( V_0 \), the energy \( E \) is:
\[ E = \frac{1}{2} C_0 V_0^2 \]

### Step 2: Analyze the first scenario (battery disconnected, plate separation doubled)
When the battery is disconnected, the charge \( Q \) on the capacitor remains constant. The capacitance when the plate separation is doubled becomes:
\[ C' = \frac{\epsilon_0 A}{2D} = \frac{C_0}{2} \]
The energy stored in the capacitor when the plate separation is doubled is:
\[ E_1 = \frac{Q^2}{2C'} = \frac{Q^2}{2 \cdot \frac{C_0}{2}} = \frac{Q^2}{C_0} = 2 \cdot \frac{1}{2} C_0 V_0^2 = 2E \]

### Step 3: Analyze the second scenario (battery connected, plate separation doubled)
When the battery remains connected, the potential \( V_0 \) remains constant. The new capacitance is still \( C' = \frac{C_0}{2} \). The energy stored in the capacitor is:
\[ E_2 = \frac{1}{2} C' V_0^2 = \frac{1}{2} \cdot \frac{C_0}{2} \cdot V_0^2 = \frac{1}{4} C_0 V_0^2 = \frac{E}{2} \]

### Step 4: Calculate the ratio \( \frac{E_1}{E_2} \)
Now we can find the ratio of the energies:
\[ \frac{E_1}{E_2} = \frac{2E}{\frac{E}{2}} = \frac{2E \cdot 2}{E} = 4 \]

### Final Answer
Thus, the ratio \( \frac{E_1}{E_2} \) is:
\[ \frac{E_1}{E_2} = 4 \]

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