A Parallel Plate capacitor is connected with a battery whose potential is constant . If the plates of capacitor are shifted apart then the intensity of electric field :

Options

A. Decrease and charge on plates also decreases.
B. Remains constant but charge on plates decreases.
C. Remains constant but charge on plates increases.
D. Increase but charge on plates decreases.

Answer

To solve the problem regarding the effect of increasing the separation between the plates of a parallel plate capacitor connected to a constant voltage battery, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Basics of a Parallel Plate Capacitor**:
- A parallel plate capacitor consists of two conductive plates separated by a distance \(D\).
- The capacitance \(C\) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{D} \]
where \(A\) is the area of the plates and \(\epsilon_0\) is the permittivity of free space.

2. **Connect the Capacitor to a Battery**:
- When connected to a battery with a constant voltage \(V\), the charge \(Q\) on the plates is given by:
\[ Q = C \cdot V = \frac{A \epsilon_0 V}{D} \]

3. **Electric Field Between the Plates**:
- The electric field \(E\) between the plates of the capacitor is defined as:
\[ E = \frac{V}{D} \]
Since the battery maintains a constant voltage \(V\), this relationship holds true.

4. **Effect of Increasing the Plate Separation \(D\)**:
- If the plates are shifted apart, the separation \(D\) increases.
- Since \(V\) is constant, the electric field \(E\) will change according to the formula:
\[ E = \frac{V}{D} \]
- As \(D\) increases, \(E\) will decrease because \(V\) remains constant.

5. **Charge on the Plates**:
- The charge \(Q\) on the plates is also affected by the increase in \(D\):
\[ Q = \frac{A \epsilon_0 V}{D} \]
- As \(D\) increases, \(Q\) decreases because it is inversely proportional to \(D\).

6. **Conclusion**:
- Therefore, when the plates of the capacitor are shifted apart, the intensity of the electric field \(E\) decreases, and the charge \(Q\) on the plates also decreases.

### Final Answer:
When the plates of a parallel plate capacitor connected to a constant voltage battery are shifted apart, the intensity of the electric field decreases.

---