Question

Obtain the formula for electrostatic force and electric pressure on the surface of the charged conductor.

Answer 1

Force on the Surface of Charged Conductor

The charge provided to a conductor is uniformly distributed on its surface. a repulsive force acts by the charge on the rest part from the the charge present at small element on the conductor, and this way, a force of repulsion acts at each small element on the conductor, and the total force on the surface of the conductor is the vector sum of force acting on all the small elements.

That is why the charged conductor experiences pressure outwards the charged conducting surface.

Let the surface charge density on the surface of the conductor be $\sigma .$ We will consider points outside and inside the conductor, two identical poimts $P_1$ and $P_2$ respectively (see fig.2.20).

Since, the electric filed outside the charged conducting surface is $\frac{\sigma }{{\epsilon }_0}$. Thus, electric field at point $P_1$

$E_{P1}=\frac{\sigma }{{\epsilon }_0}.........\left(1\right)$

Electric field inside the conductor is zero. Thus, electric field at $P_2$

$E_{P2}=\mathrm{0.........}\left(2\right)$

Now, we will divided the conductor into two parts:

(i) Element $AB$ whose surface is ds and

(ii) The remaining part $ACB$

If the electric field intensity at near point due to element $AB$ is $\overrightarrow{E_1}$ and $\overrightarrow{E_2}$ due to $ACB$ part. Then, from the fig.2.20,

$E{p}_1=E_1+E_2......\left(3\right)$

$(E_1$ and $E_2$ are in same direction at $P_1)$

and $E{p}_2=E_1-E_2.......\left(4\right)$

$(E_1$ and $E_2$ are in opposite direction at $P_2)$

From eqns. $\left(2\right)$ and $\left(4\right)$

$E_1-E_2=0$

i.e.,$E_1=E_2........\left(5\right)$

From eqns.$\left(1\right),\left(3\right)$ and $\left(5\right)$

$E_2+E_2=\frac{\sigma }{{\epsilon }_0}$

or $E_2=\frac{\sigma }{2{\epsilon }_0}$............(6)

Thus,electric field intensity at element $AB$ due to part $ACB$ is $\frac{\sigma }{2\pi {\epsilon }_0}$. If total charge on element $AB$ is $dq$, then force on element $dF=E_{2}dq=\frac{\sigma }{2{\epsilon }_0}dq(\because dq=\sigma ds)$

Thus, $dF=\frac{{\sigma }^2}{2{\epsilon }_0}ds=\frac{1}{2}{\epsilon }_0{E}^{2}ds$...........(7)$\{\because E=\frac{\sigma }{{\epsilon }_0}thus,\sigma ={\epsilon }_{0}E\}$

Force acting on whole of the surface,

$F={\oint }_S\frac{{\sigma }^2}{2{\epsilon }_0}ds={\int }_S\frac{{\epsilon }_0{E}^2}{2}ds..\left(8\right)$

$F=\frac{{\epsilon }_0{E}^2}{2}{\oint }_{S}ds$

Pressure on the unit area of surface

$P=\frac{dF}{ds}=\frac{{\sigma }^2}{2{\epsilon }_0}=\frac{1}{2}{\epsilon }_0{E}^2.......\left(9\right)$

This pressure is called electric pressure.