Question

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

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

Answer 1

Answer: (C)

$E_p=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{x}{\sqrt{R^2+x^2}}\right]$
If $x\ll R$
$E_p=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{x}{R}{\left(\frac{x^2}{R^2}+1\right)}^{-1/2}\right]$
$\Rightarrow E_p=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{x}{R}\left(1-\frac{x^2}{2R^2}\right)\right]$ (Using Binomial expansion)
$=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{x}{R}+\frac{x^3}{2R^3}\right]$
(Neglected as $x\ll R)$
Therefore, $E_p=\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{x}{R}\right)$
Here $E_p=E_p(R=b=2a)-E_p(R=a)$
$\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{h}{2a}\right)-\frac{\sigma }{2{\epsilon }_0}\left(1-\frac{h}{a}\right)]$
$=\frac{\sigma }{2{\epsilon }_0}\left(-\frac{h}{2a}+\frac{h}{a}\right)=\frac{\sigma h}{4a{\epsilon }_0}=Ch$
Thus, $C=\frac{\sigma }{4a{\epsilon }_0}$