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We have already learned in our previous discussion that field inside a conductor is zero and the field immidiately outside is

$E_{n}=\mathbf{n}\left (\frac{\sigma}{\epsilon_{0}} \right )$ (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 ?S of the surface of the conductor then force acting on area element is given by

$\Delta\mathbf{F}=\left ( \sigma \Delta S \right )\cdot \mathbf{E_{0}}$ (2)

where $\sigma$ is the surface charge density of the conductor , $\left ( \sigma \Delta S \right )$ is the amount of charge on the area element $\Delta S$ and $\mathbf{E_{0}}$ is the field in the region where charge element $\left ( \sigma \Delta S \right )$ is located.

Now there are two fields present $\mathbf{E_{\sigma}}$ and $\mathbf{E_{0}}$ and the resultant field both inside and outside the conductor near area element $\Delta S$ would be equal to the superposition of both the fields $\mathbf{E_{\sigma}}$ and $\mathbf{E_{0}}$ . Figure below shows the directions of both the fields inside and outside the conductor

_{ }on either side pointing away from the surface as shown in the figure given above. Since

**E**=0 inside the conductor

$\mathbf{E}_{in}=\mathbf{E}_{0}-\mathbf{E}_{\sigma}=0 $**
**$\mathbf{E}_{0}=\mathbf{E}_{\sigma}$

**Since direction of**

**E**

_{?}and

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

_{0}$\mathbf{E}_{out}=\mathbf{E}_{0}+\mathbf{E}_{\sigma}=2\mathbf{E}_{0}$

Thus ,

$\mathbf{E}_{0}=\mathbf{E}/2$ (3)

Equation (2) thus becomes,regardless of the of

$\Delta \mathbf{F}=\frac{1}{2}\left ( \sigma \Delta S \right )\cdot \mathbf{E}$ (4)

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

$\mathbf{f}=\frac{1}{2}\sigma \mathbf{E}$ (5)

Here is the $\textbf{E}_{\sigma }$ electric field intensity created by charge on area element $\Delta S$ at the point very close to this area element. In this region this area element behaves as infinite uniformly charged sheet hence we have,

$\textbf{E}_{\sigma }=\frac{\sigma }{2\epsilon _{0}}$ (6)

Now,

$\textbf{E}=2\textbf{E}_{0}=2\textbf{E}_{\sigma }=\frac{\sigma }{\epsilon _{0}}\hat{n}=\textbf{E}_{n}$

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

$\textbf{\textit{f}}=\frac{\sigma ^{2}}{2\epsilon _{0}} =\frac{\epsilon _{0}E^{2}}{2}\hat{n}$ (7)

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

**E**,

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

*f*