Two large metal plates (each of area $\mathbf{A}$) are held a small distance $\mathbf{d}$

a part. Suppose we put a charge${}^{\mathbf{Q}}$on each plate; what is the electrostatic pressure on the plates?

The electric field on either side of a charge-density conducting plate is as follows:

$\mathbf{E}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}{\mathbf{\epsilon }}_{\mathbf{0}}}$

Here, ${\mathit{\epsilon }}_0$is the permittivity for free space, $E$is the electric field and $\mathit{\sigma }$charge density.

The magnitude of the electric force ${}^F$acting on a charge in an electric field is expressed as,

$F=EQ$

Here,${}^Q$is the amount of the charge.

Substitute $\frac{\sigma }{2{\epsilon }_0}$for ${}^E$in above equation.

$F=\left(\frac{\sigma }{2{\epsilon }_0}\right)Q$

Write the expression for surface charge density on each plate of the capacitor is,

$\sigma =\frac{Q}{A}$

Here, $\mathrm{A}$is the area of the surface.

Substitutelocalid="1654319488470" ${}^{\frac{Q}{A}}$for ${}^{\sigma }$in the equationlocalid="1654319511537" ${}^{F=\left(\frac{\sigma }{2{\epsilon }_0}\right)Q}$

$F=\left(\frac{\left({\displaystyle \frac{Q}{A}}\right)}{2{\epsilon }_0}\right)Q$

$=\frac{Q^2}{2{\epsilon }_{0}A}$

The following formula can be used to calculate the pressure ${}^P$acting on a surface due to a force ${}^F$:

$P=\frac{F}{A}$

Substitute $\frac{Q^2}{2{\epsilon }_{0}A}$for $F$

Determine the electrostatic pressure on the parallel plate capacitor's individual plates.

role="math" localid="1654320112871" $$\begin{gathered} P=\frac{\left({\displaystyle \frac{Q^2}{2{\epsilon }_{0}A}}\right)}{A} \\ =\frac{Q^2}{2{\epsilon }_{0}A} \end{gathered}$$

Hence, the electrostatic pressure on each plate of the parallel plate capacitor is

$P=\frac{Q^2}{2{\epsilon }_0{A}^2}$