## In This post you will get answer of Q.

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

Solution:

$E_{p}=frac{sigma}{2 varepsilon_{0}}left[1-frac{x}{sqrt{R^{2}+x^{2}}}right]$

If $x ll R$

$E_{p}=frac{sigma}{2 varepsilon_{0}}left[1-frac{x}{R}left(frac{x^{2}}{R^{2}}+1right)^{-1 / 2}right]$

$Rightarrow E_{p}=frac{sigma}{2 varepsilon_{0}}left[1-frac{x}{R}left(1-frac{x^{2}}{2 R^{2}}right)right]$ (Using Binomial expansion)

$=frac{sigma}{2 varepsilon_{0}}left[1-frac{x}{R}+frac{x^{3}}{2 R^{3}}right]$

(Neglected as $x ll R)$

Therefore, $E_{p}=frac{sigma}{2 varepsilon_{0}}left(1-frac{x}{R}right)$

Here $E_{p}=E_{p}(R=b=2 a)-E_{p}(R=a)$

$left.frac{sigma}{2 varepsilon_{0}}left(1-frac{h}{2 a}right)-frac{sigma}{2 varepsilon_{0}}left(1-frac{h}{a}right)right]$

$=frac{sigma}{2 varepsilon_{0}}left(-frac{h}{2 a}+frac{h}{a}right) =frac{sigma h}{4 a varepsilon_{0}}=C h$

Thus, $C= frac{sigma}{4 a varepsilon_{0}}$