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A disc having a uniform charge density $\;\sigma\;$ and radius $\;R\;$ is placed on $\;x-y\;$ plane with centre at the origin . Find the electric potential at a point on z - axis at a distance $\;10 m\;$ from centre . $\;\sigma=16\;\in_{0}\;,R=1 m$

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

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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
$V=\large\frac{\sigma}{2\;\in_{0}}\;[x\;(1+\large\frac{R^2}{x^2})^{\large\frac{1}{2}}-x]$
$V=\large\frac{\sigma}{2\;\in_{0}}\;[x+\large\frac{R^2}{2x}-x]$
$V=\large\frac{\sigma\;R^2}{4\;\in_{0}\;x}$
$V=\large\frac{(16\;\in_{0})\;1}{4\;\in_{0}\;10}=0.4\;.$
answered Feb 4, 2014 by yamini.v
 

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