91Ó°ÊÓ

Given figure :

Theory used :

The amount of electric field that travels through a closed surface is referred to as the electric flux. The electric field through a surface is related to the charge inside the surface, according to Gauss's law, which is given by

$\mathrm{Φ}=EA$

Where $E$denotes the electric field and $A$ denotes the surface area. The electric field is always directed away from positive charges and toward negative charges.

As the box in Figure ex7 has negative charges, the net electric flux should be negative. We have three positive fields pointing out of the box:

$10N/C,10N/C,\mathrm{and}20N/C$.

Each face has the same surface area. $$\begin{gathered} {\mathrm{Φ}}_{out}=+10N/CA+10N/CA+20N/CA \\ =\underline{+40N/CA} \end{gathered}$$

is used to compute the total outward flux via the box.

$-20N/C\mathrm{and}-15N/C$are two fields that point toward the box.

Each face has the same surface area.

$$\begin{gathered} {\mathrm{Φ}}_{in}=-15N/CA-20N/CA \\ =\underline{-35N/CA} \end{gathered}$$

is used to compute the total inward flux through the box.

The net flux should be smaller than zero in total (negative)

$$\begin{gathered} {\mathrm{Φ}}_{net}<0 \\ ⇒{\mathrm{Φ}}_{out}+{\mathrm{Φ}}_{in}+{\mathrm{Φ}}_{front}<0 \\ ⇒+40N/CA-35N/CA+{\mathrm{Φ}}_{front}<0 \\ ⇒{\mathit{Φ}}_{\mathbf{f}\mathbf{r}\mathbf{o}\mathbf{n}\mathbf{t}}\mathbf{<}\mathbf{-}\mathbf{5}\mathit{N}\mathbf{/}\mathit{C}\mathit{A} \end{gathered}$$

The strength of the front electric field must be greater than $\mathbf{5}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}$and it must be inward.