A parallel-plate capacitor has charge qand plate area A.(a) By finding the work needed to increase the plate separation from xto x = dx, determine the force between the plates. (Hint:See Eq. 8-22.) (b) Then show that the force per unit area (the electrostaticstress) acting on either plate is equal to the energy density${\mathit{\epsilon }}_{\mathbf{0}}{\mathit{E}}^{\mathbf{2}}\mathbf{/}\mathbf{2}$between the plates.

Separation between the capacitor plates from x to $x+dx$.

The capacity of the capacitor to store the charge is called capacitance. It depends on the permittivity of the free space, the area of cross-section of the plates, and the separation between the two plates. When the charges are stored, the capacitor develops potential difference across the plates, due to this potential difference, capacitors also have energy stored in them. The energy can be calculated using the charge and capacitance of the capacitor.

We can find the force and force per unit area by using the formula for capacitance and energy stored in the capacitor.

Formulae:

The capacitance of the parallel capacitor,

$C=\frac{{\epsilon }_{0}A}{x}$ …(i)

Here, $C$is capacitance of the capacitor, ${\epsilon }_0$is the permittivity of the free space, $A$is area of cross section and $x$is separation between the two plates.

The energy stored in the capacitor, $U=\frac{q^2}{2C}$ …(ii)

Here,$U$ is energy stored in the capacitor,$q$ is charge, and$C$ is capacitance of the capacitor.

The force between the plates due to energy change,$F=-\frac{dU}{dx}$ …(iii)

Here,$F$ is force and$U$ is the energy stored in the capacitor.

The energy stored in the capacitor is given using equation (i) in equation (ii) as follows:

$$\begin{gathered} U=\frac{q^2}{2\left({\displaystyle \frac{{\epsilon }_{0}A}{x}}\right)} \\ =\frac{q^{2}x}{2{\epsilon }_{0}A} \end{gathered}$$

The change in energy if the separation between plates increases to$x+dx$is given by,

$dU=\frac{q^2}{2{ò}_{0}A}dx$

Thus, the force between plates can be given using the above value equation (iii) as follows:

$$\begin{gathered} F=-\frac{dU}{dx} \\ =-\frac{q^2}{2{\epsilon }_{0}A} \end{gathered}$$

The negative sign means that the force between the plates is attractive.

Hence, the value of the force is $-\frac{q^2}{2{\epsilon }_{0}A}$.

The force per unit area can be given using calculation of part (a) as follows:

$$\begin{gathered} \frac{\left|F\right|}{A}=\frac{q^2}{2{\epsilon }_0{A}^2} \\ =\frac{{\sigma }^2}{2{\epsilon }_0}(\because \sigma =\frac{q}{A}) \\ =\frac{1}{2}{\epsilon }_0{\left(\frac{\sigma }{{\epsilon }_0}\right)}^2 \\ =\frac{1}{2}{\epsilon }_0{E}^2(\because E=\frac{\sigma }{{\epsilon }_0}) \end{gathered}$$

Hence, the required value of force per area is found to be equal to the value of the energy density within the capacitors.