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

Let us have two large plates each of area A separated by a distance $d$. The charge on the plates is $\pm \mathrm{Q}$ corresponding to the charge density $\pm \sigma$.

When there is vacuum between the plates,

$\mathrm{E}_{0}=\frac{\sigma}{\epsilon_{0}}$

and the potential difference $V_{0}$.

$\therefore \mathrm{V}_{0}=\mathrm{E}_{0} d$

If $\mathrm{C}_{0}$ is the capacitance,

$\mathrm{C}_{0}=\frac{\mathrm{Q}}{\mathrm{V}_{0}}$ $=\frac{\sigma \mathrm{A}}{\mathrm{E}_{0} d} \quad[\because \mathrm{Q}=\sigma \mathrm{A}]$ $\mathrm{C}_{0}=\frac{\epsilon_{0} \mathrm{~A}}{d} \quad \ldots(1)$ $\left[\because \mathrm{E}_{0}=\frac{\sigma}{\epsilon_{0}} \Rightarrow \epsilon_{0}=\frac{\sigma}{\mathrm{E}_{0}}\right]$

Consider a dielectric inserted between the plates fully occupying the intervening region. So, the dielectric is polarised by the field and surface charge density $\pm \sigma_{\mathrm{P}}$ arise on the plates.

Hence, electric field between two plates,

$\mathrm{E}=\mathrm{E}_{0}-\mathrm{E}_{\mathrm{P}}$ $\therefore \mathrm{E}=\frac{\sigma}{\epsilon_{0}}-\frac{\sigma_{\mathrm{P}}}{\epsilon_{0}} \quad\left[\because \mathrm{E}_{\mathrm{P}}=\frac{\sigma_{\mathrm{P}}}{\epsilon_{0}}\right]$ $\therefore \mathrm{E}=\frac{\sigma-\sigma_{\mathrm{P}}}{\epsilon_{0}}$ and electric potential difference,

$\mathrm{V}=\mathrm{E} d$

$=\frac{\sigma-\sigma_{\mathrm{p}}}{\epsilon_{0}} \cdot d$

For linear dielectric $\sigma$ is proportional to $\mathrm{E}_{0}$. Hence $\sigma-\sigma_{\mathrm{P}}$ is also proportional to $\mathrm{E}$, so we can write $\sigma-\sigma_{\mathrm{P}}=\frac{\sigma}{\mathrm{K}}$

Where $K$ is a constant characteristic of dielectric.

$\therefore \mathrm{V}=\frac{\sigma d}{\epsilon_{0} \mathrm{~K}}=\frac{\mathrm{Q} d}{\mathrm{~A} \in_{0} \mathrm{~K}} \quad\left[\because \sigma=\frac{\mathrm{Q}}{\mathrm{A}}\right]$

The capacitance with dielectric between the plate is then,

$\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{A} \in_{0} \mathrm{~K}}{d}=\mathrm{K} \frac{\mathrm{A} \in_{0}}{d}$

$\therefore \mathrm{C}=\mathrm{KC}_{0}$

$\quad \ldots \text { (2) from equation (1) }$