Force on a conductor

We have already learned in our previous discussion that field inside a conductor is zero and the field immidiately outside is
E_n=n(σ/ε₀)                               (1)
where n is the unit normal vector to the surface of the conductor. We also know that any charge a conductor may carry is distributed on the surface of the conductor.
In presence of an electric field this surface charge will experience a force. If we consider a small area element ΔS of the surface of the conductor then force acting on area element is given by
ΔF=(σΔS).E₀       (2)
where σ is the surface charge density of the conductor , (σΔS) is the amount of charge on the area element ΔS and E₀ is the field in the region where charge element (σΔS) is located.
Now there are two fields present  E_σ and E₀ and the resultant field both inside and outside the conductor near area element  ΔS would be equal to the superposition of both the fields  E_σ and E₀ . Figure below shows the directions of both the fields inside and outside the conductor

Now field E₀ has same value both inside and outside the conductor and surface element ΔS suffers discontinuty because of the charge on the surface and this makes field  E_σ on either side pointing away from the surfaceas shown in the figure given above. Since E=0 inside the conductor

E<sub>in=E₀+E_σ=0
E_in=E₀=E_σ
Since direction of  E_σ and E₀ are opposite to each other and outside the conductor near its surface
E_out=E₀+E_σ=2E₀
Thus , E₀ =E/2                      (3)
Equation (2) thus becomes,regardless of the of ΔF=½(σΔS).E                  (4)
From equation 4 , force acting per unit area of the surface of the conductor is
f=½σ.E                                       (5)
Here  is the  E_σ electric field intensity created by charge on area element ΔS at the point very close to this area element. In this region this area element behaves as infinite uniformly charged sheet hence we have,
E_σ=σ/2ε₀                                                  (6)
Now,
E=2E₀=2E_σ=(σ/ε₀)n=E_n
which is in accordance with equation 1. Hence from equation 5
f=σ²/2ε₀ = (ε₀E²/2)n                (7)
This quantity f is known as surface density of force. From equation 7 we can conclude that regardless of the sign of σ and hence direction of E , f is always directed in outward direction of the conductor.