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Small elements of the conductor will experience a mechanical force normally outwards. Let us consider the small element $dS$ as shown below,

Let $\sigma $ be the surface charge density of the conductor. Then the charge carried by the small element $dS$ be $dq$.

Then we can write,

$dq = \sigma dS$ -------$(1)$

Let us consider a point $P$ outside the small element as shown in the figure.

At this point $P$, the electrical intensity can be written as,

$E = \dfrac{\sigma }{{{\varepsilon _0}k}}$ -------$(2)$

where ${\varepsilon _0}$ stands for the permittivity of free space and $k$ stands for the dielectric constant.

The electrical intensity is directed normally outwards.

The electrical intensity $E$ can be resolved into two components, ${E_1}$ and ${E_2}$.

$\overrightarrow {{E_1}} $ is the component of electrical intensity due to the small charge $dq$present in the element.

${\vec E_2}$ is the component due to the rest of the charges present on the surface.

Hence the total electrical intensity at the point $P$ can be written as,

$E = {E_1} + {E_2}$

Substituting this value in equation $(2)$

${E_1} + {E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$ --------$(3)$

Let us now consider the point $Q$. At this point, the components of electrical intensity are in opposite direction. Inside the charged conductor the total charge is zero.

i.e.

${E_1} - {E_2} = 0$

$\therefore {E_1} = {E_2}$

Substituting ${E_1} = {E_2}$ in equation $(3)$

We get

${E_2} + {E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$

This can be written as,

$2{E_2} = \dfrac{\sigma }{{{\varepsilon _0}k}}$

From this, we can write ${E_2}$ as,

${E_2} = \dfrac{\sigma }{{2{\varepsilon _0}k}}$

We know that the electrical intensity due to the rest of the charges in the conductor is ${E_2}$. The element $dS$ of the conductor will experience a repulsive force due to this component.

The repulsive force experienced by the rest of the conductor due to the charge $dq$ in the small element $dS$ is given by,

$F = {E_2}dq$

We know that ${E_2} = \dfrac{\sigma }{{2{\varepsilon _0}k}}$

and $dq = \sigma dS$

Substituting these values in the equation for force,

$F = \dfrac{\sigma }{{2{\varepsilon _0}k}}.\sigma .ds = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}dS$

From this equation, we can write

$\dfrac{F}{{dS}} = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}$

The force per unit area can be written as, $\dfrac{F}{{dS}} = f$

Thus the above equation will become,

$f = \dfrac{{{\sigma ^2}}}{{2{\varepsilon _0}k}}$

This is the mechanical force per unit area of a conductor due to a surface charge density.