Question

Force due to electrostatic pressure is directed outwards normal to the surface, .Force on a small element ds of a charged conductor
dF = (charge on ds) $\times $ Electric field $=(\sigma ds)\dfrac{\sigma }{2{{\varepsilon }_{0}}}=\dfrac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}ds$
Inside ${{E}_{1}}-{{E}_{2}}=0\Rightarrow {{E}_{1}}={{E}_{2}}$
Just outside $E={{E}_{1}}+{{E}_{2}}=2{{E}_{2}}\Rightarrow {{E}_{2}}=\dfrac{\sigma }{2{{\varepsilon }_{0}}}$
(${{E}_{1}}$ Is the field due to charge on the element $ds$ of the surface and ${{E}_{2}}$ is the field due to the rest of the sphere). The electric force acting per unit area of charged surface is defined alone.

Answer 1

Hint: The solution to this problem is obtained the by electrostatic pressure and also by using force acting on a surface .Electrostatic pressure is defined as the tension which is developed inside the conductor due to mutual repulsive force which acts among the charges of conductor and electric force acting per unit area of charged surface.

Complete step-by-step solution:
Force on the surface of charged conductor and the charge is uniformly distributed on the surface of conductor .A repulsive force acts by the charge on the rest part from the charge present at the small element on the conductor, and in this manner, a force of repulsive acts at the each small element on the conductor, and the total force acting on the surface of the conductor is the vector addition of force acting on all the small elements.
Conductor with charges experiences pressure outwards the conducting surface and the surface charge density on the conductor is $\sigma $
The electric field outside the conducting surface is $\dfrac{\sigma }{{{\varepsilon }_{0}}}$
Electric field inside the conductor is zero
Force acting on the surface,
${{F}_{s}}=\oint{s\dfrac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}}ds$
${{F}_{s}}=\oint{s\dfrac{\varepsilon {{E}^{2}}}{2}}ds$
${{F}_{s}}=\dfrac{{{\varepsilon }_{0}}{{E}^{2}}}{2}\oint{sds}$

Electrostatic pressure
In the presence of electric field surface charges on the conductor will experience a force and this force will be the average of discontinuous electric field at the surface charge .The electrostatic pressure (p) is given by
$\therefore p=\dfrac{dE}{ds}$
$\Rightarrow \dfrac{{{\sigma }^{2}}}{2{{\varepsilon }_{0}}}$
$p=\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}$

Note: Students in order to answer this question you have to derive the force equation and also electric pressure equation and at the end we get the required answer .Electrostatic pressure is obtained the differentiating the force acting on the whole surface and Electric field inside the conductor is zero. Electric pressure is called electric tension.