Question

Does the energy stored in a capacitor increase when the dielectric is removed?

Answer 1

Explanation:

We know that,

$C=k{C}_0$ ……..$\left(i\right)$

Where $k$ is the dielectric constant, $C_0$ is the capacitance with no slab between the plates, and $C$ is the capacitance of the capacitor.

And the energy stored in a capacitor can be written as:

$U=\frac{1}{2}\times \frac{Q^2}{C}$ …..$\left(ii\right)$

Where $Q$ is the charge and $C$ is the capacitance of the capacitor.

Since from the equation $\left(i\right)$ and $\left(ii\right)$ the capacitance of the capacitor is directly proportional to the dielectric constant and inversely proportional to the energy stored in a capacitor.

Therefore when the dielectric is removed, the charge on the plates, $Q$ does not change. The capacitance $C$ decreases, so the energy must increase. The capacitance decreases because the electric field increases without the polarized dielectric to cancel some of it out, or equivalently because the permittivity decreases to the vacuum permittivity.

Hence, the capacitance of the capacitor decreases when the dielectric has been removed this results in an increase in energy.