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A charge q is located at the centre of a cube. The electric flux through any face is -
A
$\dfrac{2\pi q}{6(4\pi\varepsilon_0)}$
B
$\dfrac{4\pi q}{6(4\pi\varepsilon_0)}$
C
$\dfrac{\pi q}{6(4\pi\varepsilon_0)}$
D
$\dfrac{q}{6(4\pi\varepsilon_0)}$
Detailed Solution
By Gauss's law, the total flux through the closed surface of the cube is $\dfrac{q}{\varepsilon_0}$.
The charge is at the centre, so by symmetry the flux is shared equally among the 6 faces.
Electric flux through any face $= \dfrac{\text{total flux}}{\text{number of faces}} = \dfrac{q/\varepsilon_0}{6} = \dfrac{q}{6\varepsilon_0}$
Writing this in the form of the options: $\dfrac{q}{6\varepsilon_0} = \dfrac{4\pi q}{6(4\pi\varepsilon_0)}$
So the flux through any face is $\dfrac{4\pi q}{6(4\pi\varepsilon_0)}$.
The charge is at the centre, so by symmetry the flux is shared equally among the 6 faces.
Electric flux through any face $= \dfrac{\text{total flux}}{\text{number of faces}} = \dfrac{q/\varepsilon_0}{6} = \dfrac{q}{6\varepsilon_0}$
Writing this in the form of the options: $\dfrac{q}{6\varepsilon_0} = \dfrac{4\pi q}{6(4\pi\varepsilon_0)}$
So the flux through any face is $\dfrac{4\pi q}{6(4\pi\varepsilon_0)}$.