A parallel-plate air capacitor has a capacitance of `4 muF`. What will be its new capacitance if (i) the distance between the plates is redced to half the initial distance (ii) a slab of dielectric constant 5 is introduced filling the entire space between the two plates?

Explanation

To solve the problem regarding the capacitance of a parallel-plate air capacitor, we will break it down into two parts as specified in the question.

### Given:
- Initial capacitance, \( C_1 = 4 \, \mu F \)

### Part (i): Distance between the plates is reduced to half the initial distance.

1. **Understanding Capacitance Formula**:
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