$(a)\;\large\frac{\sigma}{2 \in_{0}}\qquad(b)\;\large\frac{\sigma}{\in_{0}}\;(1-\large\frac{1}{\sqrt{\large\frac{a^2}{x^2}+1}})\qquad(c)\;\large\frac{\sigma}{2\in_{0}}\;(1-\large\frac{1}{\sqrt{\large\frac{a^2}{x^2}+1}})\qquad(d)\;zero$

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Answer : (c) $\;\large\frac{\sigma}{2\in_{0}}\;(1-\large\frac{1}{\sqrt{\large\frac{a^2}{x^2}+1}})$

Explanation :

We can assume this charge distribution as a collection of concentric rings of charge

$dA=(2 \pi r)\;dr$

$dq=\sigma\;dA=(2 \pi \sigma r)\;dr$

$dE_{x}=\large\frac{k\;(dq)\;x}{(x^2+r^2)^{\large\frac{3}{2}}}$

$=\large\frac{k\;(2 \pi \sigma r\;dr)\;x}{(x^2+r^2)^{\large\frac{3}{2}}}$

$E_{x} = \int_{0}^{a}\;dE_{x}=\int_{0}^{a}\;\large\frac{(2 \pi \sigma r k \;dr)\;x}{(x^2+r^2)^{\large\frac{3}{2}}}$

put $\;x^2+r^2=t^2$

$2r\;dr=2t\;dt$

At $\;r=0 \quad \; t=x\; : \; At \; r=a \;\quad \; t=\sqrt{x^2+a^2}$

$E_{x}=\large\frac{\sigma x}{2 \in_{0}}\;\int_{x}^{\sqrt{x^2+a^2}}\;\large\frac{t}{t^3}\;dt$

$=\large\frac{\sigma x}{2 \in_{0}}\;[-\large\frac{1}{t}]_{x}^{\sqrt{x^2+a^2}}$

$=\large\frac{\sigma x}{2 \in_{0}}\;[\large\frac{1}{x}-\large\frac{1}{\sqrt{x^2+a^2}}]$

$=\large\frac{\sigma}{2\in_{0}}\;(1-\large\frac{1}{\sqrt{\large\frac{a^2}{x^2}+1}})\;.$

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