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

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,

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

_{p}and – σ

_{p}, as shown in the figure below.

The net surface charge density then becomes equivalent to ±(σ – σ_{p}).

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 σ_{p} is proportional to E_{0 }and hence it is proportional to σ. Thus, we can say that the value (σ – σ_{p}) 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 ε_{0}K 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.

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