Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

(N/A) Consider two large plates each of area $A$ separated by a distance $d$. The charge on the plates is $\pm Q$ corresponding to the surface charge density $\pm \sigma$.
When there is a vacuum between the plates:
$E_{0} = \frac{\sigma}{\epsilon_{0}}$
The potential difference $V_{0}$ is given by:
$V_{0} = E_{0} d$
If $C_{0}$ is the capacitance:
$C_{0} = \frac{Q}{V_{0}} = \frac{\sigma A}{E_{0} d} = \frac{\epsilon_{0} A}{d} \quad \dots(1)$
When a dielectric is inserted between the plates,it becomes polarized by the external field,creating induced surface charge densities $\pm \sigma_{P}$.
The net electric field between the plates becomes:
$E = E_{0} - E_{P} = \frac{\sigma - \sigma_{P}}{\epsilon_{0}}$
The new potential difference is:
$V = E d = \frac{(\sigma - \sigma_{P}) d}{\epsilon_{0}}$
For a linear dielectric,we define the dielectric constant $K$ such that the net field is reduced by a factor $K$:
$E = \frac{E_{0}}{K} \implies \sigma - \sigma_{P} = \frac{\sigma}{K}$
Substituting this into the potential equation:
$V = \frac{\sigma d}{\epsilon_{0} K} = \frac{Q d}{A \epsilon_{0} K}$
The new capacitance $C$ is:
$C = \frac{Q}{V} = \frac{A \epsilon_{0} K}{d} = K C_{0}$
Thus,the dielectric constant $K$ is defined as $K = \frac{C}{C_{0}}$.