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Question

Eleatric field due to complete disc $(R=2 a)$ at a distance $x$ and on its axis

$E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{\sqrt{R^{2}

Vedclass · Accepted Answer

$\frac{\sigma }{{4a{ \in _0}}}$

Vedclass · Answer

Eleatric field due to complete disc $(R=2 a)$ at a distance $x$ and on its axis

$E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{x}{\sqrt{R^{2}+x^{2}}}ight]$

$E_{1}=\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{\sqrt{4 a^{2}+h^{2}}}ight]$

$ = \frac{\sigma }{{2{\varepsilon _0}}}\left[ {1 - \frac{h}{{2a}}} ight]$

$\left[ \begin{gathered}
  {	ext{ here }}x = h{	ext{ }} \hfill \
  {	ext{and }},R = 2a \hfill \ 
\end{gathered}  ight]$

Similarly, electric field due to disc $(R=a)$

$E_{2}=\frac{\sigma}{2 \varepsilon_{0}}\left(1-\frac{h}{a}ight)$

Electric field due to given disc $E=E_{1}-E_{2}$

$\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{2 a}ight]-\frac{\sigma}{2 \varepsilon_{0}}\left[1-\frac{h}{a}ight]-\frac{\sigma h}{4 \varepsilon_{0} a}$

Hence, $c=\frac{\sigma}{4 a \varepsilon_{0}}$

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