Question

Two charges of magnitudes -2Q and +Q are located at points (a,0) and (4a,0) respectively. What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?​

Answer 1

Answer:

$\phi=\frac{-Q}{18\pi a^2}$

Explanation:

As you can see in the image,

only the charge -2Q is inside the blue sphere of radius 3a. So, the flux will be

$\phi = \frac{q_{in}}{Area}=\frac{-2Q}{4\pi (3a)^2}\\\\\phi=\frac{-Q}{18\pi a^2}$

Answer 2

By Gauss' theorem, total electric flux linked with a closed surface is given by Ф = q/ ε₀ where, q is the total charge enclosed by the closed surface.

∴ Total electric flux linked with cube, Ф = q/ ε₀

As charge is at centre, therefore, electric flux is symmetrically distributed through all 6 faces.

Flux linked with each face = 1/6Ф = (1/6) × (q/ ε₀) = q/6ε₀