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Distribution of charges is shown in figure . The flux of electric field due to these charges through the surface is

$(a)\;\large\frac{q}{\in_{0}}\qquad(b)\;zero\qquad(c)\;\large\frac{2q}{\in_{0}}\qquad(d)\;\large\frac{3q}{\in_{0}}$

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Answer : (d) $\;\large\frac{3q}{\in_{0}}$
Explanation :
$\phi=\large\frac{q_{enc}}{\in_{0}}\;.$
$q_{enc}=q+2q=3q$
$\phi=\large\frac{3q}{\in_{0}}\;.$

 

answered Jan 30, 2014 by yamini.v
edited Aug 11, 2014 by thagee.vedartham
 

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