**E**

_{n}=

**n**(σ/ε

_{0}) (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

Δ

**F**=(σΔS)

**.E**

_{0}(2)

where σ is the surface charge density of the conductor , (σΔS) is the amount of charge on the area element ΔS and

**E**

_{0}is the field in the region where charge element (σΔS) is located.

Now there are two fields present

**E**

_{σ}and

**E**

_{0}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

**E**

_{0}. Figure below shows the directions of both the fields inside and outside the conductor

Now field E

_{0}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=

**E**

_{0}+

**E**

_{σ}=0

**E**

_{in}=

**E**

_{0}=

**E**

_{σ}

Since direction of

**E**

_{σ}and

**E**are opposite to each other and outside the conductor near its surface

_{0}**E**

_{out}=

**E**

_{0}+

**E**

_{σ}=2

**E**

_{0}

Thus ,

**E**

_{0}=

**E/**2 (3)

Equation (2) thus becomes,regardless of the of Δ

**F**=½(σΔS)

**.E**(4)

From equation 4 , force acting per unit area of the surface of the conductor is

**=½σ**

*f***.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/2ε

_{0}(6)

Now,

**E**=2

**E**

_{0}=2

**E**

_{σ}=(σ/ε

_{0})

**n**=

**E**

_{n}

which is in accordance with equation 1. Hence from equation 5

**=σ**

*f*^{2}/2ε

_{0}= (ε

_{0}E

^{2}/2

**)n**(7)

This quantity

**is known as surface density of force. From equation 7 we can conclude that regardless of the sign of σ and hence direction of**

*f***E**,

**is always directed in outward direction of the conductor.**

*f*
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