Question

Derive an expression for mechanical force per unit area of surface charge density.

Answer 1

Hint: The quantity of charge contained per unit area is called surface charge density. All charged conductors will experience some mechanical force. Elements having a surface charge density will always experience a mechanical force outwards. Here we have to derive an expression for the mechanical force.

Complete step by step answer:
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,
$E={\displaystyle \frac{\sigma }{{\epsilon }_{0}k}}$ -------$(2)$
where ${\epsilon }_0$ 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.
${\overrightarrow{E}}_2$ 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 $(2)$
$E_1+E_2={\displaystyle \frac{\sigma }{{\epsilon }_{0}k}}$ --------$(3)$

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
$E_2+E_2={\displaystyle \frac{\sigma }{{\epsilon }_{0}k}}$
This can be written as,
$2E_2={\displaystyle \frac{\sigma }{{\epsilon }_{0}k}}$
From this, we can write $E_2$ as,
$E_2={\displaystyle \frac{\sigma }{2{\epsilon }_{0}k}}$
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 $E_2={\displaystyle \frac{\sigma }{2{\epsilon }_{0}k}}$
and $dq=\sigma dS$
Substituting these values in the equation for force,
$F={\displaystyle \frac{\sigma }{2{\epsilon }_{0}k}}.\sigma .ds={\displaystyle \frac{{\sigma }^2}{2{\epsilon }_{0}k}}dS$
From this equation, we can write
${\displaystyle \frac{F}{dS}}={\displaystyle \frac{{\sigma }^2}{2{\epsilon }_{0}k}}$
The force per unit area can be written as, ${\displaystyle \frac{F}{dS}}=f$
Thus the above equation will become,
$f={\displaystyle \frac{{\sigma }^2}{2{\epsilon }_{0}k}}$
This is the mechanical force per unit area of a conductor due to a surface charge density.

Note: The S.I unit for the mechanical force per unit area is given by $N/m^2$. Like charges repel each other and unlike charges attract. In a charge-carrying conductor, the charges are all similar. Hence when we consider a small element the rest of the conductor will experience the mechanical force due to the repulsion of like charges.