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# A point charge Q is located on the axis of a disc is radius R at a distance b from the plane of the disc . Find the electric flux through the disc if R= $\;\sqrt{3}\;b$

$(a)\;\large\frac{Q}{4\;\in_{0}}\qquad(b)\;\large\frac{Q}{2\;\in_{0}}\qquad(c)\;\large\frac{Q}{3\;\in_{0}}\qquad(d)\;\large\frac{Q}{8\;\in_{0}}$

Answer : (a) $\;\large\frac{Q}{4\;\in_{0}}$
Explanation : $\;E_{x}\;$ on the ring of radius x $\;=\large\frac{k\;Q}{(b^2+x^2)}\;cos\;\theta$
$E_{x}=\large\frac{k\;Q\;b}{(b^2+x^2)^{\large\frac{3}{2}}}$
$\phi=\oint\;E_{x}\;.da$
$\phi=\int_{x=0}^{x=R}\;\large\frac{k\;Q\;b\;2\;\pi\;x\;dx}{(b^2+x^2)^{\large\frac{3}{2}}}$
Putting $\;x=b\;tan\;\theta$
$\phi=\large\frac{2\;\pi\;k\;Q\;b}{b^3}\;\int_{0}^{tan^{-1}\;(\large\frac{R}{b})}\;\large\frac{(b\;tan\;\theta)\;(b\;sec^{2}\;\theta)}{sec^3\;\theta}\;d \theta$
$\phi=2\;\pi\;k\;Q \int_{0}^{\alpha}\;sin\;\theta\;d \theta\quad\;[\alpha=tan^{-1} (\sqrt{3})=60^{0}]$
$\phi=2\;\pi\;k\;Q\;(1-cos \alpha)$
$\phi=\large\frac{Q}{2\;\in_{0}}\times\large\frac{1}{2}=\large\frac{Q}{4\;\in_{0}}\;.$