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Find the electric field caused by a disc of radius "a" with a uniform charge density $\;\sigma\;$ , at a point along the axis of the disc a distance x from its centre .

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

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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
$dA=(2 \pi r)\;dr$
$dq=\sigma\;dA=(2 \pi \sigma r)\;dr$
$=\large\frac{k\;(2 \pi \sigma r\;dr)\;x}{(x^2+r^2)^{\large\frac{3}{2}}}$
$E_{x} = \int_{0}^{a}\;dE_{x}=\int_{0}^{a}\;\large\frac{(2 \pi \sigma r k \;dr)\;x}{(x^2+r^2)^{\large\frac{3}{2}}}$
put $\;x^2+r^2=t^2$
At $\;r=0 \quad \; t=x\; : \; At \; r=a \;\quad \; t=\sqrt{x^2+a^2}$
$E_{x}=\large\frac{\sigma x}{2 \in_{0}}\;\int_{x}^{\sqrt{x^2+a^2}}\;\large\frac{t}{t^3}\;dt$
$=\large\frac{\sigma x}{2 \in_{0}}\;[-\large\frac{1}{t}]_{x}^{\sqrt{x^2+a^2}}$
$=\large\frac{\sigma x}{2 \in_{0}}\;[\large\frac{1}{x}-\large\frac{1}{\sqrt{x^2+a^2}}]$
answered Feb 12, 2014 by yamini.v
edited Feb 12, 2014 by yamini.v

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