Question

A thin disc of radius b=2a has a concentric hole of radius 'a' is in it. It carries uniform surface charge ${}^{\prime }{\sigma }^{\prime }$ on it. If the electric field on its axis at height 'h' (h<<a) from its centre is given as 'Ch' the value of 'C' is.

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

Answer 1

Answer: (D)

The electric field due to uniform charge disk with surface charge density $\sigma$ is given as

$E=\frac{\sigma }{2ϵ}[1-\frac{h}{\sqrt{h^2+R^2}}]$ where h is distance along the axis of disk from center of disk

The electric field at distance h from center for given distribution can be found by subtracting the electric field due to disk of radius a from the disk of radius 2b

$E_{net}=E_{diskradius2a}-E_{diskradiusa}$

$E=\frac{\sigma }{2ϵ}(1-\frac{h}{\sqrt{h^2+(2a{)}^2}})-\frac{\sigma }{2ϵ}(1-\frac{h}{\sqrt{h^2+a^2}})$..(i)

as $h<<a\sqrt{h^2+(2a{)}^2}\simeq 2a$ and $\sqrt{h^2+(a{)}^2}\simeq a$ substituting in equation (i)

$E=\frac{\sigma }{2ϵ}(1-\frac{h}{2a})-\frac{\sigma }{2ϵ}(1-\frac{h}{a})$

$E=\frac{\sigma }{2ϵ}\frac{h}{2a}$

$E=\frac{\sigma h}{4ϵa}$

$E=\frac{\sigma }{4ϵa}h$

The value of C is $C=\frac{\sigma }{4ϵa}$ hence option C is correct