Question

You connect a $\mathbf{\text{9 V}}$battery to a capacitor consisting of two circular plates of radius $\mathbf{\text{0.08 m}}$separated by an air gap of $\mathbf{\text{2mm}}$, what is the charge on the positive plate?

Short answer

The charge on the capacitor is $0.8\text{nC}$.

Step-by-step solution

A concept:

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 circular plates,$r=0.08\text{m}$

Voltage, $V=9\text{V}$

The width of the gap between the plates is,

$\begin{array}{rcl}s& =& 2\text{mm}\\ & =& \left(2\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ & =& 2\times {10}^{-3}\text{m}\end{array}$

Define capacitance:

The capacitance of a parallel plate capacitor given by a relation,

$C=\frac{{\epsilon }_{0}A}{s}$

Here, $s$is the width of the gap between the plates, $A$ is the area of the capacitor’s plate, and ${\epsilon }_0$is the permittivity of free space having a value localid="1662143076540" $8.85\times {10}^{-12}\raisebox{1ex}{${\text{C}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^{\text{2}}$}\right.$.

The relation between capacitor and the charge on the plates of the capacitor can be expressed as,

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

Here, $Q$is the charge on the plates and $V$ is the potential between the plates.

By comparing equations (1) and (2), you have

$$\begin{gathered} \frac{{\epsilon }_{0}A}{s}=\frac{Q}{V} \\ Q=\frac{{\epsilon }_{0}AV}{s} \end{gathered}$$

Since the plates are in a circular shape, the area of plates is taken as$\pi r^2$. Thus,

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

Here, $r$ is the radius of circular plates.

Substitute known values in the above equation.

localid="1668576415138" $\begin{array}{rcl}Q& =& \frac{\left(8.85\times {10}^{-12}{\displaystyle \raisebox{1ex}{${\text{C}}^{\text{2}}$}\!\left/ \!\raisebox{-1ex}{$\text{N}·{\text{m}}^{\text{2}}$}\right.}\right)\times 3.14\times {\left(0.08\text{m}\right)}^2\times \left(9\text{V}\right)}{2\times {10}^{-3}\text{m}}\\ & =& 0.8\times {10}^{-9}\text{C}\\ & =& 0.8\text{nC}\end{array}$

Hence, the charge on the capacitor is $0.8\text{nC}$.