What is the electric flux through each of the surfaces A to E in FIGURE Q24.6? Give each answer as a multiple of $\frac{q}{{\epsilon }_0}$.

Given Figure :

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, that is :

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

(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 total contained charge in figure (A) is$Q_{in}=\Sigma Q=\left(3q\right)+\left(q\right)=\underline{4q}$

As a result, the electric flux is

${\Phi }_A=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{4}\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

(B) Two charges are in figure (B).

As a result,$Q_{in}=\Sigma Q=(-3q)+(-q)=\underline{-4q}$equals the total enclosed charge.

As a result, the electric flux is equal to${\Phi }_B=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{-}\mathbf{4}\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

(C) Two charges are in figure (C) .

As a result, $Q_{in}=\Sigma Q=\left(q\right)+(-q)=\underline{0}$

As a result, the electric flux is equal to ${\Phi }_C=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{0}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

(D) Three charges are in figure (D).

As a result, $Q_{in}=\Sigma Q=\left(3q\right)+\left(q\right)+(-q)=\underline{3q}$

As a result, the electric flux is equal to ${\Phi }_D=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{3}\mathit{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$

(E) Four charges are in figure (E).

As a result, $Q_{in}=\Sigma Q=\left(3q\right)+\left(q\right)+(-3q)+(-q)=\underline{0}$

As a result, the electric flux is equal to ${\Phi }_E=\frac{Q_{in}}{{\epsilon }_0}=\boxed{\frac{\mathbf{0}}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$