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A charge Q is placed at the corner of a cube electric flux through the cube is

$(a)\;\large\frac{Q}{6 \in_{0}}\qquad(b)\;\large\frac{Q}{8 \in_{0}}\qquad(c)\;\large\frac{Q}{\in_{0}}\qquad(d)\;\large\frac{Q}{3\in_{0}}$

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Answer : (b) $\;\large\frac{Q}{8\in_{0}}$
Explanation :
Only $\;\large\frac{1}{8}^{th}\;$ part of total flux will pass through this cube . therefore
$\phi_{cube}=\large\frac{\phi_{total}}{8}=\large\frac{Q}{8 \in_{0}}\;.$
answered Jan 30, 2014 by yamini.v
 

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