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A charge $q$ is located at the centre of a cube. The electric flux through any face is
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Verified Answer
The correct answer is:
$\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}$
According to Gauss's theorem, electric flux through the cube (closed surface),
$\phi=\frac{q}{\varepsilon_0}$
Since, cube has six surfaces and all the faces are symmetrical, therefore electric flux through any face
$\phi^{\prime}=\frac{q}{6 \varepsilon_0}=\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}$
$\phi=\frac{q}{\varepsilon_0}$
Since, cube has six surfaces and all the faces are symmetrical, therefore electric flux through any face
$\phi^{\prime}=\frac{q}{6 \varepsilon_0}=\frac{4 \pi q}{6\left(4 \pi \varepsilon_0\right)}$
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