Question

On introducing dielectric in a capacitor:
(i) Electric field decreases
(ii) Capacitance increases
(iii) Potential increases
(iv) Capacitance decreases.

A. i, ii, iii are true.
B. i, ii, iv are true.
C. only iii and iv are true.
D. only i and ii are true.

Answer 1

Hint: A capacitor is a circuit element which is used to store energy in the circuit. It stores energy in the form of an electric field. Capacitor is a passive component of the circuit. More such examples of passive components are resistors and inductors. Dielectric constant is a value which may be defined as the ratio of intensity of electric field in vacuum to that in the dielectric medium.

Formula used:
 $\kappa = \dfrac{E_\circ}{E},\ C = \dfrac{\epsilon_{\circ}A}{d},\ V = -\int E.dr \ or \ V = Ed$

Complete answer:
Dielectric constant is denoted by ’$\kappa$’ (Greek - kappa). It may be defined as the property of a material to diminish the interaction between the charges. Since dielectric constant is the ratio of electric field in free space to that in the medium, so greater the dielectric constant value, lesser will be the value of electric field inside the medium. Dielectric constant of air (or free space) is 1. Hence, if we insert a dielectric medium in the capacitor, the electric field decreases. Now, since $V = Ed$, hence potential also decreases. Now, on inserting a dielectric material having dielectric constant ‘$\kappa$’, the capacitance increases $\kappa$times. Thus if initially the capacitance were $C = \dfrac{\epsilon_{\circ}A}{d}$, on inserting the slab, it becomes $C = \dfrac{\kappa \epsilon_{\circ}A}{d}$.

So, the correct answer is “Option D.

Note:
On introducing the di-electric medium between two charges, the interaction between the charges gets decreased by a factor called di-electric constant of the medium. Every medium has their own dielectric constant which defines its electric properties. Water has the maximum value of dielectric constant i.e. $\kappa = 81$. Students should remember that $\kappa > 1$.