A charge q is placed at one corner of a cube | vedclass

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(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q}{\varepsilon_{0}}$.Since the charge $q$ is placed at one cor

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$\frac{q}{24\varepsilon_{0}}$

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(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q}{\varepsilon_{0}}$.Since the charge $q$ is placed at one corner of the cube,it is shared by $8$ such identical cubes to enclose the charge completely.Therefore,the total flux through the cube is $\phi_{	ext{cube}} = \frac{q}{8\varepsilon_{0}}$.The flux through the three faces of the cube that meet at the corner where the charge is placed is zero because the electric field lines are parallel to these surfaces.The remaining three faces are symmetric with respect to the charge,so the flux through each of these three faces is equal.Let the shaded area be one of these faces. The flux through the shaded area is $\phi_{	ext{shaded}} = \frac{1}{3} 	imes \phi_{	ext{cube}} = \frac{1}{3} 	imes \frac{q}{8\varepsilon_{0}} = \frac{q}{24\varepsilon_{0}}$.

$A$ charge $q$ is placed at one corner of a cube as shown in the figure. The flux of the electrostatic field $\overrightarrow{E}$ through the shaded area is ...... .

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