A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then:

Options

A. some charge from the capacitor will flow back into the source.
B. some extra charge from the source will flow back into the capacitor.
C. the electric field intensity between the two plate does not change.
D. the electric field intensity between the two plates will decrease.

Answer

To solve the question regarding the effects of introducing a dielectric plate into a parallel plate air-core capacitor connected to a constant potential difference, we can analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Initial State of the Capacitor:**
- A parallel plate capacitor consists of two plates separated by a distance \(d\) and filled with air (or vacuum).
- The capacitance \(C\) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \(A\) is the area of the plates and \(\varepsilon_0\) is the permittivity of free space.

2. **Charge on the Capacitor:**
- When connected to a constant potential difference \(V\), the charge \(Q\) on the capacitor is:
\[ Q_i = C V = \frac{\varepsilon_0 A}{d} V \]

3. **Introducing the Dielectric:**
- When a dielectric material with dielectric constant \(K\) is introduced between the plates, the new capacitance \(C'\) becomes:
\[ C' = \frac{K \varepsilon_0 A}{d} \]
- The dielectric increases the capacitance because it reduces the electric field between the plates.

4. **Final Charge on the Capacitor:**
- The new charge \(Q_f\) on the capacitor after the dielectric is introduced can be calculated as:
\[ Q_f = C' V = \frac{K \varepsilon_0 A}{d} V \]
- Since \(C'\) is greater than \(C\), \(Q_f\) will be greater than \(Q_i\).

5. **Effect on Charge Flow:**
- Because the capacitor is connected to a source of constant potential difference, when the dielectric is introduced, some extra charge will flow from the source to the capacitor to accommodate the increased capacitance. Thus:
\[ Q_f > Q_i \]
- This means that extra charge flows from the source into the capacitor.

6. **Electric Field Intensity:**
- The electric field \(E\) between the plates of the capacitor is given by:
\[ E = \frac{V}{d} \]
- Since the potential difference \(V\) remains constant and the distance \(d\) does not change, the electric field intensity remains the same when the dielectric is introduced.

7. **Conclusion:**
- Therefore, the correct statements regarding the effects of introducing a dielectric into the capacitor are:
- Some extra charge from the source will flow into the capacitor.
- The electric field intensity between the two plates does not change.

### Summary of Answers:
- **Correct Statements:**
- Some extra charge from the source will flow back into the capacitor.
- The electric field intensity between the two plates does not change.