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

Answer 1

Answer: (B)

Option (b): both V and E decrease

Answer 2

For an isolated capacitor, the charge $Q$ remains constant. The initial capacitance is $C$ and voltage is $V$, so:

$$Q = C V.$$

When a dielectric with dielectric constant $\kappa$ is inserted, the new capacitance becomes:

$$C' = \kappa C.$$

Since $Q$ remains constant, the new voltage $V'$ is:

$$V' = \frac{Q}{C'} = \frac{C V}{\kappa C} = \frac{V}{\kappa}.$$

Thus, the voltage decreases.

The energy stored in a capacitor is given by:

$$E = \frac{1}{2} C V^2.$$

After inserting the dielectric, the new energy $E'$ is:

$$E' = \frac{1}{2} C' V'^2 = \frac{1}{2} (\kappa C) \left(\frac{V}{\kappa}\right)^2 = \frac{1}{2} \kappa C \frac{V^2}{\kappa^2} = \frac{1}{2} \frac{C V^2}{\kappa} = \frac{E}{\kappa}.$$

So, the energy also decreases.