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

Hint: If a differential area on the surface of the conductor, and calculate the net electric field of the differential area and the rest of the conductor. Inside a charged conductor, the total electric field is zero.

Formula used: In this solution we will be using the following formulae;
 $ E = \dfrac{\sigma }{\varepsilon } $
 where $ E $ is the electric field due to a closed charged surface at a point outside the closed charged surface, $ \sigma $ is the surface charge density of the surface, and $ \varepsilon $ is the permittivity of free space.
 $ dF = Edq $ where $ dF $ is an infinitesimal force exerted by an infinitesimal charge $ dq $ .
 $ q = \sigma A $ where $ q $ is the charge on the surface $ A $ is the area of a surface.
 $ P = \dfrac{{dF}}{{dA}} $ where $ P $ is the pressure acting on an infinitesimal surface due to an infinitesimal force.

Complete step by step answer:
Generally, we know that when the surface of a conductor is charged, there are repelling forces which the charges exert on each other. And the charges being on the conductor causes an outward force on the conductor tending to increase its volume.
Now to calculate the force, we pick a differential element of the surface $ dA $ with an electric field $ {E_1} $ . We let the rest of the surface have an electric field $ {E_2} $ . Generally, we know that the electric field outside a closed charged surface is given as $ E = \dfrac{\sigma }{\varepsilon } $ where $ E $ is the electric field due to a closed charged surface at a point outside the closed charged surface, $ \sigma $ is the surface charge density of the surface, and $ \varepsilon $ is the permittivity of free space.
This electric field is the sum of the electric field of the differential surface and that of the rest of the conductor. Hence,
 $ {E_1} + {E_2} = \dfrac{\sigma }{\varepsilon } $
Also, we know that the electric field inside a closed charged surface is zero. Hence choosing a point close to the differential are but inside the charged surface we have that
 $ {E_1} - {E_2} = 0 $
 $ \Rightarrow {E_1} = {E_2} $
Hence, we have
 $ {E_1} + {E_1} = \dfrac{\sigma }{\varepsilon } $
 $ \Rightarrow 2{E_1} = \dfrac{\sigma }{\varepsilon } $
Hence, we have
 $ {E_1} = \dfrac{\sigma }{{2\varepsilon }} $
The force exerted by an infinitesimal charge is given by
 $ dF = Edq $ where $ dF $ is an infinitesimal force exerted by an infinitesimal charge $ dq $ .
Hence,
 $ d{F_1} = {E_1}dq $
But from $ q = \sigma A $ where $ q $ is the charge on the surface $ A $ is the area of a surface, we have $ dq = \sigma dA $ , also $ E = \dfrac{\sigma }{\varepsilon } $
Then,
 $ d{F_1} = \dfrac{{{\sigma ^2}}}{{2\varepsilon }}dA $
Then the force on the entire surface is the sum of the tiny differential area of the surface. We use integral as in
 $ F = \oint_S {d{F_1}} = \oint_S {\dfrac{{{\sigma ^2}}}{{2\varepsilon }}dA} $
 $ \Rightarrow F = \dfrac{{{\sigma ^2}}}{\varepsilon }\oint_S {dA} $
For the pressure, we have $ P = \dfrac{{dF}}{{dA}} $ .
The pressure of a uniform distribution of charge is also uniform, hence we say $ P = \dfrac{{d{F_1}}}{{dA}} $ which from above will give $ P = \dfrac{{\dfrac{{{\sigma ^2}}}{{2\varepsilon }}dA}}{{dA}} = \dfrac{{{\sigma ^2}}}{{2\varepsilon }} $ .

Note:
For clarity, the fact that the sum of the electric field for the point inside the conductor is given by $ {E_1} - {E_2} = 0 $ is because, inside the conductor, the electric field will be in the opposite direction. Basically it should be $ {E_1} + \left( { - {E_2}} \right) = 0 $ .