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A charge $q$ is placed at the corner of a cube of side $a$. The electric flux through the cube is
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Verified Answer
The correct answer is:
$\frac{q}{8 \varepsilon_{0}}$
According to Gauss's law, the electric flux through a closed surface is equal to $\frac{1}{\varepsilon_{0}}$ times the net charge enclosed by the surface. Since, $q$ is the charge enclosed by the surface, then the electric flux $\phi=\frac{q}{\varepsilon_{0}}$. If charge $q$ is placed at a corner of cube, it will be divided into 8 such cubes. Therefore, electric flux through the cube is
$$
\phi^{\prime}=\frac{1}{8}\left(\frac{q}{\varepsilon_{0}}\right)
$$
$$
\phi^{\prime}=\frac{1}{8}\left(\frac{q}{\varepsilon_{0}}\right)
$$
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