# 2.16    ( a) Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by $(E_{1}-E_{2}).\widehat{n}= \frac{\sigma }{\epsilon _{0}}$ where nˆ is a unit vector normal to the surface at a point and σ is the surface charge density at that point.  Hence, show that just outside a conductor, the electric field is $\sigma \frac{\widehat{n}}{\epsilon _{0}}$

Answers (1)

The electric field on one side of Surface with charge density $\sigma$

$E_1=-\frac{\sigma}{2\epsilon _0}\widehat{n}$

The electric field on another side of Surface with charge density $\sigma$

$E_2=-\frac{\sigma}{2\epsilon _0}\widehat{n}$

Now, resultant of both surfaces:

As E1 and E2 are opposite in direction. we have

$E_1-E_2=\frac{\sigma}{2\epsilon _0}-\left ( -\frac{\sigma}{2\epsilon _0} \right )\widehat{n}=\frac{\sigma}{\epsilon _0}$

There has to be a discontinuity at the sheet of the charge since both electric fields are in the opposite direction.

Now,

Since the electric field is zero inside the conductor,

the electric field just outside the conductor is

$E=\frac{\sigma}{\epsilon _0}\widehat{n}$

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