A square of side 20 cm is enclosed by a sphe | vedclass

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(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$,where $q_{

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(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$,where $q_{enclosed}$ is the net charge enclosed by the surface.Since the square is placed inside the spherical surface and all four charges are at the corners of the square,all these charges are enclosed by the spherical surface.The net charge $q_{enclosed} = (2 - 5 - 3 + 6) 	imes 10^{-6} \ C = 0 	imes 10^{-6} \ C = 0 \ C$.Therefore,the total flux $\phi = \frac{0}{\epsilon_0} = 0 \ N \cdot m^2/C$.

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