Question

How does the energy stored in a capacitor change when a dielectric is inserted if (a) the capacitor is isolated so Q does not change; (b) the capacitor remains connected to a battery so V does not change? Explain.

Short answer

(a) The value of energy stored reduces when a dielectric is inserted and the capacitor is isolated.

(b) The value of energy stored increases when a dielectric is inserted and the capacitor remains connected to a battery.

Step-by-step solution

Understanding the dielectric

Dielectric is a material that has zero free electrons and is a poor conductor of electricity.

(a) Evaluation of the change in energy stored when capacitor is isolated

The relation of energy stored in a capacitor is given by,

\(E = \frac{1}{2}\frac{{{Q^2}}}{C}\)

Here, Eis the energy stored, Cis the capacitance and Qis the charge.

When the dielectric is added, the relation of energy stored will become,

\(E = \frac{1}{2}\frac{{{Q^2}}}{{KC}}\)

Here, Kis the dielectric constant.

The value of dielectric constant is always \(K > 1\). From the above relation, it is seen that the value of energy stored reduces by the addition of dielectric.

(b) Evaluation of the change in energy stored when capacitor remains connected to a battery

The relation of energy stored in a capacitor is given by,

\(E = \frac{1}{2}C{V^2}\)

Here, Eis the energy stored, Cis the capacitance and Vis the voltage.

When the dielectric is added, the relation of energy stored will become,

\(E = \frac{1}{2}KC{V^2}\)

Here, Kis the dielectric constant and its value is always\(K > 1\).

In the above relation, it is seen that the value of energy stored increases by the addition of dielectric