Question

Two circular plates of radius $\mathbf{\text{0.12 m}}$are separated by an air gap of $\mathbf{\text{1.5 mm}}$. The plates carry charge$\mathbf{+}\mathit{Q}$and$\mathbf{-}\mathit{Q}$where $\mathit{Q}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{\text{C}}$. (a) What is the magnitude of the electric field in the gap? (b) What is the potential difference across the gap? (c) What is the capacitance of this capacitor?

Short answer

(a) The electric field is $8.99\times {10}^{-12}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{C}$}\right.$.

(b) The potential difference between the plates is $135\text{V}$.

(c) The capacitance of the capacitor is $2.66\times {10}^{-10}\text{F}$.

Step-by-step solution

A concept:

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.

Given data:

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

The charge, $Q=3.6\times {10}^{-8}\text{C}$

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

(a) The magnitude of the electric field in the gap:

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

$E=\frac{Q}{A{\epsilon }_0}$

Here, $Q$is the charge, $A$is the area of the capacitor, ${\epsilon }_0$and is the permittivity of free space having a value of $8.85\times {10}^{-12}\raisebox{1ex}{${\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" $\pi r^2$.

Thus, the electric field will be,

$E=\frac{Q}{\left(\pi r^2\right){\epsilon }_0}$

Substitute known values in the above equation.

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

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

(b) The potential difference across the gap:

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.

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

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

(c) The capacitance of the capacitor:

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

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

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

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

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