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.

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

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.

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