On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of `in_(r)=10` is introduced between the plates, without disconnecting the d.c. source.
Explain, using suitable expressions, how the (i) capacitance, (ii) electric field and (iii) electric field and (iii) energy density of the capacitor change.

Explanation

To solve the problem step by step, we will analyze how the capacitance, electric field, and energy density of a parallel plate capacitor change when the spacing between the plates is halved and a dielectric medium is introduced, while keeping the DC source connected.

### Step 1: Capacitance Calculation
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
where:
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