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Effect Of Dielectric On Capacitance

Effect Of Dielectric On Capacitance :- Suppose the charge densities on the plates of a capacitor are +σ and −σ respectively, and the distance between the plates is d. The entire space between the plates is filled with a dielectric material having dielectric constant ε_r.

The dielectric material is polarized by the external electric field , resulting in surface charge densities of + and – on its surfaces. Therefore, the total (resultant) surface charge density on the plates will be .

Therefore, the resultant electric field in the dielectric material between the plates :

$\displaystyle E=\frac{\sigma -{{\sigma }_{p}}}{{{\varepsilon }_{0}}}$

Now the potential difference between the plates :

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

Now we know that the induced charge density, σ_p:

$\displaystyle {{\sigma }_{p}}=\sigma \left( 1-\frac{\text{1}}{{{\varepsilon }_{r}}} \right)$

Therefore

$\displaystyle {{\sigma }_{p}}=\sigma -\frac{\sigma }{{{\varepsilon }_{r}}}$

$\displaystyle \Rightarrow \sigma -{{\sigma }_{p}}=\frac{\sigma }{{{\varepsilon }_{r}}}$

Again the potential difference between the plates :

$\displaystyle V=\frac{\sigma d}{{{\varepsilon }_{0}}{{\varepsilon }_{r}}}$

$\displaystyle \Rightarrow V=\frac{Qd}{A{{\varepsilon }_{0}}{{\varepsilon }_{r}}}$

Capacitance :

$\displaystyle C=\frac{Q}{V}=\frac{Q}{\frac{Qd}{A{{\varepsilon }_{0}}{{\varepsilon }_{r}}}}={{\varepsilon }_{r}}\left( \frac{{{\varepsilon }_{0}}A}{d} \right)$