A charge 'Q mu C' is placed at the centre of | vedclass

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(A) According to Gauss's Law,the total electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{q_{enclosed}}{\epsilon_{0}}$.For a cube

Vedclass · Accepted Answer

$\frac{Q}{6 \epsilon_{0}} \mu Vm, \quad \frac{Q}{3 \epsilon_{0}} \mu Vm$

Vedclass · Answer

(A) According to Gauss's Law,the total electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{q_{enclosed}}{\epsilon_{0}}$.For a cube with a charge $Q$ placed at its centre,the total flux through all six faces is $\Phi_{total} = \frac{Q}{\epsilon_{0}}$.Since the cube is symmetric,the flux through each of the six faces is equal.Flux through one face: $\phi_{one} = \frac{\Phi_{total}}{6} = \frac{Q}{6 \epsilon_{0}}$.Flux through two opposite faces: $\phi_{two} = 2 	imes \phi_{one} = 2 	imes \frac{Q}{6 \epsilon_{0}} = \frac{Q}{3 \epsilon_{0}}$.Therefore,the flux through one face and two opposite faces is $\frac{Q}{6 \epsilon_{0}}$ and $\frac{Q}{3 \epsilon_{0}}$ respectively.

$A$ charge '$Q$ $\mu C$' is placed at the centre of a cube. The flux through one face and two opposite faces of the cube is respectively:

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