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If the electric flux entering and leaving an enclosed surface is $\phi_{1}\;and\;\phi_{2}\;.$ The charge inside the surface will be

$(a)\;(\phi_{2}-\phi_{1})\;\in_{0}\qquad(b)\;(\phi_{1}+\phi_{2})\;\in_{0}\qquad(c)\;\large\frac{(\phi_{2}-\phi_{1})}{ \in_{0}}\qquad(d)\;\large\frac{(\phi_{1}+\phi_{2})}{\in_{0}}$

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Answer : (a) $\;(\phi_{2}-\phi_{1})\;\in_{0}$
Explanation : Net flux through surface = $\;\phi_{2}-\phi_{1}$
Using gauss's law
$\phi_{2}-\phi_{1}=\large\frac{q_{enc}}{\in_{0}}$
$q_{enc}=(\phi_{2}-\phi_{1})\;\in_{0}\;.$
answered Feb 3, 2014 by yamini.v
 

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