Question

A parallel plate air capacitor has capacity 'C' farad, potential 'V' volt and energy 'E' joule. When the gap between the plates is completely filled with dielectric

• A. V increases, E decreases.
• B. V decreases, E increases.
• C. both V and E increase.
• D. both V and E decrease.

Hint:

For an isolated charged capacitor the charge stays constant, so a larger capacitance lowers both V and E.

Answer 1

The Correct Option is D

Step 1: Understand the concept
The capacitor is charged and then isolated, so the charge \(Q\) on its plates cannot change when the dielectric is inserted. Filling the gap with a dielectric of constant \(K > 1\) raises the capacitance to \(C' = KC\).

Step 2: Effect on potential
\[ V' = \frac{Q}{C'} = \frac{Q}{KC} = \frac{V}{K} \]
So the potential decreases.

Step 3: Effect on energy
\[ E' = \frac{Q^2}{2C'} = \frac{E}{K} \]
The stored energy also decreases, as the plates must do work on the dielectric as it is pulled in.

Step 4: Result
Both \(V\) and \(E\) decrease, option (D). Options (A), (B) and (C) each have at least one of the two quantities increasing, which would need \(C'\) to be less than \(C\).

Final Answer:
Both V and E decrease. This is option (D). \[ \boxed{\text{(D) }\text{Both V and E decrease}} \]