Question

Charges $q_1=-4Q\mathrm{and}{q}_2=+2Q$are located at$x=-a\mathrm{and}x=+a$, respectively. What is the net electric flux through a sphere of radius $2a$centered

(a) at the origin and

(b) at $x=2a$?

Short answer

a. ${\Phi }_e=\boxed{\frac{\mathbf{-}\mathbf{2}\mathbf{Q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

b.${\Phi }_e=\boxed{\frac{\mathbf{2}\mathbf{Q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

Step-by-step solution

Given information and Theory used 

Given : Charges : $q_1=-4Q\mathrm{and}{q}_2=+2Q$

Located at : $x=-a\mathrm{and}x=+a$, respectively.

Radius of sphere : $2a$

Theory used :

The quantity of electric field that passes through a closed surface is referred to as the electric flux. The electric flux through a surface is proportional to the charge inside the surface, according to Gauss's law, which is given by :

${\Phi }_e=\oint \overrightarrow{E}·d\overrightarrow{A}=\frac{Q_{in}}{{\epsilon }_0}$ (1)

Calculating the net electric flux through the sphere centered at the origin

(a) The electric flow is determined by the charge inside the closed surface, as indicated. There is no flux owing to charges outside the closed surface.

The enclosed charges are $-4Q\mathrm{and}+2Q$when the sphere is centered at the origin, as indicated in the diagram below, hence the enclosed charge $Q_{in}$in the sphere is the sum of both charges.

$Q_{in}=-4Q+2Q=-2Q$

To derive the flux, we plug the values for $Q_{in}$into equation (1).

${\Phi }_e=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{-}\mathbf{2}\mathbf{Q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

Calculating the net electric flux through the sphere centered at x=2a

(b) When the sphere is centered at $x=2a$as shown below, the enclosed charge is just $+2Q$, so the enclosed charge is $Q_{in}=+2Q$in the sphere.

To derive the flux, we plug the values for $Q_{in}$into equation (1).

${\Phi }_e=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{2}\mathbf{Q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$