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. $\frac{\sigma}{a\epsilon_{0}}$
• B. $\frac{\sigma}{2a\epsilon_{0}}$
• C. $\frac{\sigma}{4a\epsilon_{0}}$
• D. $\frac{\sigma}{8a\epsilon_{0}}$

Answer 1

Answer: (C)

$E_{p}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{\sqrt{R^{2}+x^{2}}}\right]$
If $x \ll R$
$E_{p}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{R}\left(\frac{x^{2}}{R^{2}}+1\right)^{-1 / 2}\right]$
$\Rightarrow E_{p}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{R}\left(1-\frac{x^{2}}{2 R^{2}}\right)\right]$ (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, $C= \frac{\sigma}{4 a \varepsilon_{0}}$