91Ó°ÊÓ

A capacitor stores potential energy in its electric field. This energy is proportional to the charge on the plates and the area of the plates.

A parallel plate capacitor is an arrangement of two metal plates connected in parallel and separated from each other by a certain distance. The gap between the plates is occupied by a dielectric medium.

The radius of both the plats, $r=0.12\text{m}$

The charge, $Q=3.6×{10}^{-8}\text{C}$

Distance between two plates, $d=1.5\text{mm}=1.5×{10}^{-3}\text{m}$

The electric field between the plates of the parallel plate capacitor can be expressed as,

$E=\frac{Q}{A{\mathrm{Î\mu }}_0}$

Here, $Q$is the charge, $A$is the area of the capacitor, ${\mathrm{Î\mu }}_0$and is the permittivity of free space having a value of $8.85×{10}^{-12}\raisebox{1ex}{${\text{C}}^2$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^2$}\right.$.

Since the plates are in circular shape, the area of the plates is replaced by localid="1662195341045" $\mathrm{Ï€}{r}^2$.

Thus, the electric field will be,

$E=\frac{Q}{\left(\mathrm{Ï€}{r}^2\right){\mathrm{Î\mu }}_0}$

Substitute known values in the above equation.

$\begin{array}{rcl}E& =& \frac{3.6×{10}^{-8}\text{C}}{3.14×{\left(0.12\text{m}\right)}^2×8.85×{10}^{-12}{\displaystyle \raisebox{1ex}{${\text{C}}^2$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^2$}\right.}}\\ & =& 8.99×{10}^{-12}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.\end{array}$

Hence, the electric field is localid="1662197042428" $8.99×{10}^{-12}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.$.

The potential difference between plates of the capacitor is equal to the product of the electric field between the plates and the separation between the plates of the capacitor.

$V=Ed$

Here, $E$is the electric field and $d$is the separation between the plates.

Substitute known values in the above equation.

$\begin{array}{rcl}V& =& \left(8.99×{10}^{-12}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.\right)×\left(1.5×{10}^{-3}\text{m}\right)\\ & =& 135\text{V}\end{array}$

Hence, the potential difference between the plates is $135\text{V}$.

The capacitance of the capacitor in terms of charge $Q$and potential $V$can be expressed as,

$C=\frac{Q}{V}$

Substitute $3.6×{10}^{-8}\text{C}$for $Q$and $135\text{V}$for $V$in the above equation.

role="math" localid="1662196657118" $\begin{array}{rcl}C& =& \frac{3.6×{10}^{-8}\text{C}}{135\text{V}}\\ & =& 2.66×{10}^{-10}\text{F}\end{array}$

Hence, the capacitance of the capacitor is $2.66×{10}^{-10}\text{F}$.