Question

The electric flux (in SI units) through any face of a cube due to a positive charge \(Q\) situated at the centre of the cube is

• A. \(\dfrac{Q}{2\varepsilon_0}\)
• B. \(4\pi\varepsilon_0 Q\)
• C. \(\dfrac{Q}{6\varepsilon_0}\)
• D. \(\dfrac{Q}{\varepsilon_0}\)
• E. \(6\pi\varepsilon_0 Q\)

Hint:

If a charge is placed at the centre of a symmetric closed surface, flux splits equally across identical faces.

Answer 1

The Correct Option is C

By Gauss's law, total flux through the cube is: \[ \Phi = \frac{Q}{\varepsilon_0} \] Since the charge is at the centre, symmetry tells us that flux is equally distributed through all \(6\) faces. Thus flux through one face: \[ \Phi_{\text{one face}}=\frac{1}{6}\cdot \frac{Q}{\varepsilon_0} =\frac{Q}{6\varepsilon_0} \] Hence, \[ \boxed{(C)\ \frac{Q}{6\varepsilon_0}} \]