Question

A very thin disc is uniformly charged with surface charge density $\sigma >0$. Then the electric field intensity on the axis at the point from which the disc is seen at an solid angle $\Omega$ is

• A. $\frac{\sigma \Omega }{2\pi {ϵ}_0}$
• B. $\frac{\sigma \Omega }{4\pi {ϵ}_0}$
• C. $\frac{\sigma \Omega }{\pi {ϵ}_0}$
• D. $\frac{\sigma \Omega }{8\pi {ϵ}_0}$

Answer 1

The correct option is A $\frac{\sigma \Omega }{2\pi {ϵ}_0}$

$\sigma >0$

Electric field intensity on the axis at the point from which $=\Omega$ angle.

Electric field near on infinite plane of uniform charge density.

A much more important limit of the above result is actually for $x$ much less than $P$. In this case, it is as through the disk were of infinite extent. So, the result correspond simply to the electric field near on infinite shell of these.

$E_x=2\pi K\sigma$

$=\frac{\sigma \Omega 2\pi }{4\pi {ϵ}_0}K=\frac{\Omega }{4\pi {ϵ}_0}$

$=2\Omega \frac{\sigma \Omega }{2\pi {ϵ}_0}$