Question

A charge $Q \, \mu $ C is placed at the centre of cube, the flux coming out from any surface will be :

• A. $ \frac{Q}{24\varepsilon_{0}} $
• B. $ \frac{Q}{8\varepsilon _{0}} $
• C. $ \frac{Q}{6\varepsilon _{0}}\times 10^{-6} $
• D. $ \frac{Q}{6\varepsilon _{0}}\times 10^{-3} $

Answer 1

Answer: (C)

From Gauss theorem, the electric flux through any closed surface is equal to $\frac{1}{\varepsilon_{0}}$ times the net charge $Q$ enclosed by the surface, that is,
$\phi_{E}=\int E \cdot d A=\frac{q}{\varepsilon_{0}}$
Total flux through six faces of the cube is

$\phi=\frac{Q \times 10^{-6}}{\varepsilon_{0}}$
Since, cube has six faces, therefore flux through one face is
$\phi_{s}=\frac{1 \times Q \times 10^{-6}}{6 \varepsilon_{0}}$
$\phi_{s}=\frac{Q \times 10^{-6}}{6 \varepsilon_{0}} \frac{ Nm ^{2}}{ C }$