$(a)\;0.1\qquad(b)\;0.4\qquad(c)\;1.6\qquad(d)\;None\;of\;these$

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Answer : (b) $\;0.4$

Explanation : potential due to whole disc = $\;\large\frac{\sigma}{2\;\in_{0}}\;[\sqrt{R^2+x^2}-x]$

Since x > > R

$V=\large\frac{\sigma}{2\;\in_{0}}\;[x\;(1+\large\frac{R^2}{x^2})^{\large\frac{1}{2}}-x]$

$V=\large\frac{\sigma}{2\;\in_{0}}\;[x+\large\frac{R^2}{2x}-x]$

$V=\large\frac{\sigma\;R^2}{4\;\in_{0}\;x}$

$V=\large\frac{(16\;\in_{0})\;1}{4\;\in_{0}\;10}=0.4\;.$

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