A point charge q is placed at a distance a/2 | vedclass

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(D) Consider the square of side $a$ as one face of a cube of side $a$. Since the charge $q$ is placed at a distance $a/2$ from the center of the squar

Vedclass · Accepted Answer

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

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(D) Consider the square of side $a$ as one face of a cube of side $a$. Since the charge $q$ is placed at a distance $a/2$ from the center of the square,it is located at the geometric center of this imaginary cube. According to Gauss's Law,the total electric flux $\Phi_{	ext{total}}$ through the closed surface of the cube is given by $\Phi_{	ext{total}} = \frac{q}{\epsilon_0}$. Since the cube has $6$ identical faces and the charge is placed symmetrically at the center,the flux through each face is equal. Therefore,the electric flux $\Phi$ through the square face is $\Phi = \frac{\Phi_{	ext{total}}}{6} = \frac{q}{6 \epsilon_0}$.

$A$ point charge $q$ is placed at a distance $a/2$ directly above the center of a square of side $a$. The electric flux through the square is:

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