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

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

\(E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{\sqrt{4 a^{2}+h^{2}}}\right]\)

\( = \frac{\sigma }{{2{\varepsilon _0}}}\left[ {1 - \frac{h}{{2a}}} \right]\)

\(\left[ \begin{gathered}
  {\text{ here }}x = h{\text{ }} \hfill \\
  {\text{and }},R = 2a \hfill \\ 
\end{gathered}  \right]\)

Similarly, electric field due to disc \((R=a)\)

\(E_{2}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{h}{a}\right)\)

Electric field due to given disc \(E=E_{1}-E_{2}\)

\(\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{2 a}\right]-\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{a}\right]-\frac{\sigma h}{4 \varepsilon_{0} a}\)

Hence, \(c=\frac{\sigma}{4 a \varepsilon_{0}}\)