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A point charge Q is placed at the centre of an imaginary Gaussian surface. Find your through curved surface.


$(A)\;\frac{al}{2 \in _0 \sqrt {\frac{l^2}{4} +R^2}} \\ (B)\;zero \\ (C)\; \frac{Q}{2 \in_0 }\bigg[ 1- \frac{l}{2 \sqrt {R^2+\frac{l^2}{4} }}\bigg] \\ (D)\;\frac{Q}{\in_0 }\bigg[ 1- \frac{l}{2 \sqrt {R^2+\frac{l^2}{4} }}\bigg] $

1 Answer

$\phi =\phi _{curved}+\phi _{circle_1}+\phi _{circle 2}$
=> $\phi =\phi _{cs}+2 \phi _{c1} \bigg[ \phi _{c_1}=\phi _{c_2}\bigg]$
=> $\phi _{curved}= \phi _{cs}=\phi -2 \phi _{c_1}$
$\qquad= \large\frac{Q}{\in _0}$$- 2 \phi _{c_1}$
$\overrightarrow{E} =\large\frac{Q}{4 \pi \in_0} \times \frac{1}{(x^2+l^2/4)}$
$d \phi _{c1}=\overrightarrow {E} . d \overrightarrow{s} =E \times 2\pi x dx \times \cos \theta$
$\qquad= \large\frac{Q}{2 \in _0} \times \frac{x dx}{(x^2 +l^2/4)} \times \frac{l/2}{\sqrt {x^2 +l^2/4}}$
$\qquad= \large\frac{Ql}{4 \in _0} \frac{x dx}{(x^2 +l^2/4)^{3/2}} $
=> $ \phi _{c1}= \large\frac{Ql}{4 \in _0} \int \limits_0^R \frac{x dx}{(x^2 +l^2/4)^{3/2}}$
$\phi_{cs}$= $\large\frac{Q}{2 \in_0 }\bigg[ 1- \frac{l}{2 \bigg[R^2+\frac{l^2}{4} \bigg]^{1/2}}\bigg]$
$=>\large \frac{Q}{2 \in_0 }\bigg[ 1- \frac{l}{2 \sqrt {R^2+\frac{l^2}{4} }}\bigg]$
Hence C is the correct answer.


answered Jan 13, 2014 by meena.p
edited Aug 1, 2014 by thagee.vedartham

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