91Ó°ÊÓ

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 :

${\mathrm{Φ}}_e=âˆ\circledR \stackrel{→}{E}·d\stackrel{→}{A}=\frac{Q_{in}}{{\mathrm{Î\mu }}_0}$ (1)

(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).

${\mathrm{Φ}}_e=\frac{Q_{in}}{{\mathrm{Î\mu }}_0}=\boxed{\frac{\mathbf{-}\mathbf{2}\mathbf{Q}}{{\mathbf{Î\mu }}_{\mathbf{0}}}}$

(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).

${\mathrm{Φ}}_e=\frac{Q_{in}}{{\mathrm{Î\mu }}_0}=\boxed{\frac{\mathbf{2}\mathbf{Q}}{{\mathbf{Î\mu }}_{\mathbf{0}}}}$