If a charge Q is placed at the corner of a cu | vedclass

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(C) According to Gauss's Law,the total flux through a closed surface is $\phi_{total} = \frac{Q}{\varepsilon_0}$.When a charge $Q$ is placed at the co

Vedclass · Accepted Answer

$\frac{Q}{24\varepsilon_0}$

Vedclass · Answer

(C) According to Gauss's Law,the total flux through a closed surface is $\phi_{total} = \frac{Q}{\varepsilon_0}$.When a charge $Q$ is placed at the corner of a cube,it is shared by $8$ such identical cubes to enclose the charge completely.Therefore,the flux passing through the entire cube is $\phi_{cube} = \frac{1}{8} \left( \frac{Q}{\varepsilon_0} ight) = \frac{Q}{8\varepsilon_0}$.The charge is placed at a corner,so the electric field lines are parallel to the three faces meeting at that corner. Thus,the flux through these three faces is zero.The remaining $3$ faces share the total flux equally.Therefore,the flux through one of these faces is $\phi_{face} = \frac{1}{3} 	imes \phi_{cube} = \frac{1}{3} 	imes \frac{Q}{8\varepsilon_0} = \frac{Q}{24\varepsilon_0}$.

If a charge $Q$ is placed at the corner of a cube,what is the electric flux passing through one of its faces?

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