A parallel plate air capacitor has initial capacitance C. If plate separation is slowly increased from `d_(1)` to `d_(2)`, then mark the correct statement (s). (Take potential of the capacitor to be constant, i.e. throughout the process it remains connected to battery.)

Options

A. Work done by electric force=negative of work done by external agent.
B. Work done by external force`=-intoverset(vec)F.overset(vec)dx,` where `overset(vec)(F)` is the electric force of attraction between the plates at plate setaration `x`.
C. Work done by electric force `ne`negative of work done by external agent.
D. Work done by battery `=` two times the change in electric potential energy stored in capacitor.

Explanation

To solve the problem, we need to analyze the behavior of a parallel plate capacitor when the plate separation is increased while it remains connected to a battery, keeping the potential constant.

### Step-by-Step Solution:

1. **Understanding Capacitance**:
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon A}{d} \]
where \( \varepsilon \) is the permittivity of the medium (air in this case), \( A \) is the area of the plates, and \( d \) is the separation between the plates.

2. **Initial and Final Capacitance**:
- Initial capacitance when the separation is \( d_1 \):
\[ C_1 = \frac{\varepsilon A}{d_1} \]
- Final capacitance when the separation is \( d_2 \):
\[ C_2 = \frac{\varepsilon A}{d_2} \]

3. **Charge on the Capacitor**:
Since the capacitor is connected to a battery, the voltage \( V \) across the capacitor remains constant. The charge \( Q \) on the capacitor is given by:
\[ Q = C \cdot V \]
- Initial charge:
\[ Q_1 = C_1 \cdot V = \frac{\varepsilon A}{d_1} \cdot V \]
- Final charge:
\[ Q_2 = C_2 \cdot V = \frac{\varepsilon A}{d_2} \cdot V \]

4. **Change in Charge**:
The change in charge as the separation increases from \( d_1 \) to \( d_2 \) is:
\[ \Delta Q = Q_2 - Q_1 = \left(\frac{\varepsilon A}{d_2} - \frac{\varepsilon A}{d_1}\right) V \]

5. **Work Done by the Battery**:
The work done by the battery when the charge changes is given by:
\[ W = \Delta Q \cdot V \]
Substituting for \( \Delta Q \):
\[ W = \left(\frac{\varepsilon A}{d_2} - \frac{\varepsilon A}{d_1}\right) V^2 \]

6. **Change in Electric Potential Energy**:
The electric potential energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
- Initial energy:
\[ U_1 = \frac{1}{2} \cdot C_1 \cdot V^2 = \frac{1}{2} \cdot \frac{\varepsilon A}{d_1} \cdot V^2 \]
- Final energy:
\[ U_2 = \frac{1}{2} \cdot C_2 \cdot V^2 = \frac{1}{2} \cdot \frac{\varepsilon A}{d_2} \cdot V^2 \]
- Change in energy:
\[ \Delta U = U_2 - U_1 = \frac{1}{2} \left(\frac{\varepsilon A}{d_2} - \frac{\varepsilon A}{d_1}\right) V^2 \]

7. **Comparing Work Done and Change in Energy**:
From the expressions for work done by the battery and change in potential energy, we can conclude:
\[ W = 2 \Delta U \]
Therefore, the work done by the battery is equal to twice the change in electric potential energy stored in the capacitor.

### Conclusion:
Based on the analysis:
- **Option A**: Work done by electric force is equal to negative of work done by external agent. **(Correct)**
- **Option B**: Work done by external agent is equal to minus integration \( f \cdot dx \). **(Correct)**
- **Option C**: Work done by electric force is not equal to negative of work done by external agent. **(Incorrect)**
- **Option D**: Work done is equal to 2 times the change in electric potential energy stored in capacitor. **(Correct)**