A parallel plate air condenser is charged and then disconnected from the charging battery. Now the space between the plates is filled with a dielectric then, it electric field strength between the plates

Options

A. increases while its capacity increases
B. increases while its capacity decreases
C. decreases while its capacity increases
D. decreases while its capacity decreases

Answer

To solve the problem step by step, we need to analyze the situation of a parallel plate capacitor that has been charged, disconnected from the battery, and then filled with a dielectric material.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- A parallel plate capacitor is charged with charge \( Q \) and voltage \( V \). The capacitance \( C \) of the capacitor is given by the formula:
\[ C = \frac{Q}{V} \]

2. **Disconnecting from the Battery**:
- Once the capacitor is charged and disconnected from the battery, the charge \( Q \) remains constant on the plates of the capacitor.

3. **Filling with Dielectric**:
- When the space between the plates is filled with a dielectric material, the dielectric constant \( k \) comes into play. The new capacitance \( C' \) of the capacitor with the dielectric is given by:
\[ C' = kC \]
- This means the capacitance increases when the dielectric is introduced.

4. **Electric Field Calculation**:
- The electric field \( E \) between the plates of a capacitor is given by:
\[ E = \frac{V}{d} \]
- Where \( d \) is the separation between the plates. Since the charge \( Q \) is constant, the voltage across the capacitor changes when the dielectric is introduced.

5. **Finding the New Voltage**:
- The new voltage \( V' \) across the capacitor with the dielectric can be expressed in terms of the charge and the new capacitance:
\[ V' = \frac{Q}{C'} = \frac{Q}{kC} \]
- Since \( C = \frac{Q}{V} \), we can substitute to find:
\[ V' = \frac{Q}{k \frac{Q}{V}} = \frac{V}{k} \]

6. **New Electric Field**:
- Now, substituting \( V' \) back into the electric field equation gives:
\[ E' = \frac{V'}{d} = \frac{V/k}{d} = \frac{E}{k} \]
- This shows that the electric field strength decreases when the dielectric is introduced.

7. **Conclusion**:
- Therefore, the electric field strength between the plates decreases when the dielectric is introduced. The final electric field strength is:
\[ E' = \frac{E}{k} \]

### Final Answer:
The electric field strength between the plates decreases when the dielectric is introduced.