Find the electric flux through the shaded fac | vedclass

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Question

(C) The total electric flux through the face $BCGF$ is the sum of the fluxes due to individual charges.
Using Gauss's Law and symmetry arguments

Vedclass · Accepted Answer

$\frac{q}{6\epsilon_0}$

Vedclass · Answer

(C) The total electric flux through the face $BCGF$ is the sum of the fluxes due to individual charges.
Using Gauss's Law and symmetry arguments for a cube:
$\phi_{BCGF} = \phi_{	ext{due to } q} + \phi_{	ext{due to } 3q} + \phi_{	ext{due to } 2q}$
For a charge placed at a corner of a cube,the flux through each of the three adjacent faces is $0$,and through each of the three opposite faces is $\frac{q}{24\epsilon_0}$.
For charge $q$ at corner $E$,the face $BCGF$ is an opposite face,so $\phi_{	ext{due to } q} = \frac{q}{24\epsilon_0}$.
For charge $3q$ at corner $A$,the face $BCGF$ is an opposite face,so $\phi_{	ext{due to } 3q} = \frac{3q}{24\epsilon_0}$.
Since the charge $2q$ lies on the plane of the face $BCGF$,the electric field lines are parallel to the surface,and the flux through this face due to $2q$ is $0$.
Therefore,$\phi_{BCGF} = \frac{q}{24\epsilon_0} + \frac{3q}{24\epsilon_0} + 0 = \frac{4q}{24\epsilon_0} = \frac{q}{6\epsilon_0}$.

Find the electric flux through the shaded face $BCGF$ of the cube,as shown in the figure.

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