Question

A parallel plate capacitor is charged by a battery and the battery remains connected, a dielectric slab is inserted in the space between the plates. Explain what changes if any, occur in the values of the
i) Potential difference between the plates
ii) Electric field between the plates
iii) Energy stored in the capacitor

Answer 1

Hint: The capacitance of a capacitor depends on its physical dimensions and the material between its plates. The capacitance increases when a dielectric is placed in the space between its plates. The energy of the capacitor depends directly on the capacitance and the electric field depends inversely on it. Since the battery remains connected, the potential difference has to remain constant for the same potential drop in the circuit.

Formula used:
Capacitance $C$ of a capacitor of plate area $A$ and distance between plates $d$ is given by
$C=\dfrac{AK{{\varepsilon }_{0}}}{d}$
where $K$ is the dielectric constant of the material between the plates and ${{\varepsilon }_{0}}$ is the permittivity of free space equal to ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$. For air $K=1$.
Electric field $E$ between the plates of a capacitor is given by
$E=\dfrac{Q}{AK{{\varepsilon }_{0}}}$
where $Q$ is the magnitude of charge on a plate of the capacitor with air between its plates.
The energy of the capacitor is given by
$\text{Energy = }\dfrac{1}{2}C{{V}^{2}}$

Complete step by step answer:
(i) When a dielectric is introduced inside the space between the plates of a capacitor, while the battery is still connected, the potential difference between both the plates has to remain constant, so that the algebraic sum of the potential drop in the circuit remains constant.
The dielectric gets polarized in a direction opposite to the pre-existing electric field of the capacitor and thus, produces an opposing electric field of its own, which reduces the electric field of the capacitor. Hence, the plates can accumulate more charge and the capacitance can increase, since capacitance is nothing but a measure of the amount of charge that can be stored by the capacitor when a potential difference is applied across its plates.
Capacitance $C$ of a capacitor of plate area $A$ and distance between plates $d$ is given by
$C=\dfrac{AK{{\varepsilon }_{0}}}{d}$ --(1)
where $K$ is the dielectric constant of the material between the plates and ${{\varepsilon }_{0}}$ is the permittivity of free space equal to ${{\varepsilon }_{0}}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$. For air $K=1$.
Hence, when a dielectric is introduced into the capacitor, its capacitance increases by a factor of $K$, that is, the dielectric constant of the introduced dielectric.

(ii) Electric field $E$ between the plates of a capacitor is given by
$E=\dfrac{Q}{AK{{\varepsilon }_{0}}}$ --(2)
where $Q$ is the magnitude of charge on a plate of the capacitor with air between its plates.
Now, from (2), it can be clearly seen that the electric field decreases by a factor of $\dfrac{1}{K}$ when the dielectric is introduced between the plates.

(iii) The energy of the capacitor is given by
$\text{Energy = }\dfrac{1}{2}C{{V}^{2}}$ --(3)
Now, since from (1), we can see that the capacitance increases by a factor of $K$, upon the introduction of the dielectric, since the potential difference $V$ remains constant as explained earlier, from (3), we can see that the energy of the capacitor will also increase by a factor of $K$.

Note: Students must keep in mind that since in the question, it is mentioned that the battery remains connected and the dielectric is introduced, the potential difference remains constant. However, if the capacitor was isolated and the same process was done, the problem would have to be solved by considering that the charge on the plates of the capacitor would remain constant and hence, the problem would change completely. Then in fact, the energy of the capacitor would decrease and not increase by a factor of $K$, though the capacitance would still increase. This is because the potential difference across the plates would decrease.