Assertion. Capacity of a parallel plate condenser remains unaffected on introduced a conducting or insulating slab between the plates.
Reason. In both the cases, electric field intensity between the plates increases.

Options

A. If both assertion and reason are ture and reason is the correct explanation of assertion.
B. If both assertin and reason are ture but reason is not the correct explanation of assertion .
C. If assertion is true but reason is false.
D. If both assertion and reason are false.

Explanation

To solve the problem, we need to analyze the assertion and the reason provided in the question regarding the capacity of a parallel plate capacitor when a conducting or insulating slab is introduced between the plates.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that the capacity (capacitance) of a parallel plate capacitor remains unaffected when a conducting or insulating slab is introduced between the plates.

2. **Understanding the Reason**:
- The reason given is that in both cases (with a conducting or insulating slab), the electric field intensity between the plates increases.

3. **Analyzing the Effect of a Conducting Slab**:
- When a conducting slab is placed between the plates of a capacitor, it divides the capacitor into two capacitors in series. The electric field inside a conductor is zero, which means that the effective distance between the plates is reduced.
- If the original distance between the plates is \(D\), and the thickness of the conducting slab is \(d\), the new distance \(D' = D - d\).
- The capacitance \(C\) of a parallel plate capacitor is given by:
\[ C = \frac{\varepsilon_0 A}{D} \]
- The new capacitance \(C'\) becomes:
\[ C' = \frac{\varepsilon_0 A}{D'} \]
- Since \(D' < D\), it follows that \(C' > C\). Thus, the capacitance increases with the introduction of a conducting slab.

4. **Analyzing the Effect of an Insulating Slab**:
- When an insulating (dielectric) slab is introduced, it reduces the electric field between the plates. The electric field \(E\) in the presence of a dielectric is given by:
\[ E' = \frac{E_0}{k} \]
- Here, \(k\) is the dielectric constant (greater than 1). The capacitance with the dielectric becomes:
\[ C' = \frac{k \varepsilon_0 A}{D} \]
- Since \(k > 1\), it follows that \(C' > C\). Therefore, the capacitance also increases with the introduction of an insulating slab.

5. **Conclusion**:
- Both the assertion and the reason are false. The capacity of a parallel plate capacitor is affected by the introduction of either a conducting or insulating slab, and the electric field intensity actually decreases in both cases.

### Final Answer:
- The assertion is false, and the reason is also false.