Effect of Dielectric on Capacitance

We have read about the working of a parallel plate capacitor and know that the value of capacitance is dependent on the material or the medium between the two plates. In this section, we will learn in details, how the capacitance of a parallel plate capacitor changes when a dielectric medium is inserted between its plates.

If we consider a simple capacitor with parallel plates of area A, separated by a distance d, we can see that for a capacitor with charge Q, the charge on each plate is +Q and –Q. As the area of the plate is A, the corresponding charge density can be given as ±σ. Where,

\(\sigma =\frac{Q}{A}\)

When the two plates have vacuum between them, potential energy across the capacitor can be given as,

\(V_{0}=E_{0}d=V_{0}\frac{\sigma }{\varepsilon _{0}}d\)

The capacitance of the capacitor can thus be given as,

\(C_{0}=\frac{Q}{V_{0}}=\varepsilon _{0}\frac{A}{d}\)

Let us consider another capacitor with the same specifications as taken before. Let us insert a dielectric between the plates such that it fully occupies the space between the plates. As the dielectric enters the field between the plates, it gets polarized by the field and the charges get arranged such that they act as two charged sheets with a surface charge density of σp and – σp, as shown in the figure below.

The net surface charge density then becomes equivalent to ±(σ – σp).

The potential energy across the capacitor can thus be given as,

\(V=Ed=\frac{\sigma -\sigma _{p}}{\varepsilon _{0}}d\)

In case of linear dielectrics, we can say that σp is proportional to E0 and hence it is proportional to σ. Thus, we can say that the value (σ – σp) is also proportional to σ. Mathematically,

\(\sigma -\sigma _{p}=\frac{\sigma }{K}\)

Where K is a constant whose value depends upon the dielectric medium selected. The potential energy across the capacitor can this be written as,

\(V=\frac{\sigma d}{\varepsilon _{0}K}=\frac{Qd}{A\varepsilon _{0}K}\)

And the capacitance between the plates can be given as,

\(C=\frac{Q}{V}=\frac{\varepsilon _{0}KA}{d}\)

Here ε0K is the permittivity of the medium, which can also be given as,

\(\varepsilon =\varepsilon _{0}K\)

Here the value K is the permittivity of the medium such that, for a given medium,

\(K=\frac{\varepsilon }{\varepsilon _{0}}\)

And the ratio of the capacitance of the capacitor with a dielectric medium to the capacitor with vacuum between the plates can be given as,

\(K=\frac{C}{C _{0}}\)

Stay tuned with Byju’s to learn more about the effect of dielectric on a capacitor and other related topics.


Practise This Question

Find the effective capacitance of the following system. Having a variable dielectric