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

Electric filed due to complete disc (R=2a) at a a distance x and on its axis
$\begin{array}{l}{\mathrm{E}}_1=\frac{\mathrm{\sigma }}{2{\mathrm{ϵ}}_0}\left[1-\frac{\mathrm{x}}{\sqrt{{\mathrm{R}}^2+{\mathrm{x}}^2}}\right]{\mathrm{E}}_1=\frac{\mathrm{\sigma }}{2{\in }_0}\left[1-\frac{\mathrm{h}}{\sqrt{4{\mathrm{a}}^2+{\mathrm{h}}^2}}\right]\\ =\frac{\mathrm{\sigma }}{2{\mathrm{ϵ}}_0}\left[1-\frac{\mathrm{h}}{2\mathrm{a}}\right]\left[\begin{array}{c}\text{ here }\mathrm{x}=\mathrm{h}\\ \text{ and }\mathrm{R}=2\mathrm{a}\end{array}\right]\end{array}$
Similarly ,electric field due to disc (R =a)
${\mathrm{E}}_2=\frac{\mathrm{\sigma }}{2{\in }_0}\left[1-\frac{\mathrm{h}}{\mathrm{a}}\right]$
Electric field due to given disc
$\begin{array}{l}\mathrm{E}={\mathrm{E}}_1-{\mathrm{E}}_2\\ \frac{\mathrm{\sigma }}{2{\mathrm{ϵ}}_0}\left[1-\frac{\mathrm{h}}{2\mathrm{a}}\right]-\frac{\mathrm{\sigma }}{2{\mathrm{ϵ}}_0}\left[1-\frac{\mathrm{h}}{\mathrm{a}}\right]=\frac{\mathrm{\sigma h}}{4{\mathrm{ϵ}}_0\mathrm{a}}\end{array}$
Hence, $\mathrm{c}=\frac{\mathrm{\sigma }}{4{\mathrm{aϵ}}_0}$