Question

What is the electric flux through each of the surfaces in FIGURE Q24.5? Give each answer as a multiple of $\frac{q}{{\epsilon }_0}$.

Short answer

In figure (a),the electric flux is $\frac{\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

In figure (b),the electric flux is$\frac{\mathbf{-}\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

In figure (c),the electric flux is$\mathbf{0}$

Step-by-step solution

Given information and theory used

Given Figures :

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 :

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

Finding the electric flux through surface (a)

(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

$$\begin{gathered} Q_{in}=\Sigma Q=\left(q\right)+\left(q\right)+\left(q\right)-\left(q\right)-\left(q\right) \\ =\underline{q} \end{gathered}$$

As a result, the electric flux is

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

Finding the electric flux through surface (b)

(b) Two positive charges are outside the enclosed surface in figure (b), while two negative charges and one positive charge are inside the contained charge.

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

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

Finding the electric flux through surface (c)

(c) All charges are outside the enclosed surface in figure (c). As a result, the total enclosed charge is $Q_{in}=\underline{0}$.

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