91Ó°ÊÓ

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

$$\begin{gathered} {\mathrm{Φ}}_{\mathrm{e}}=âˆ\circledR \stackrel{→}{E}×d\stackrel{→}{A} \\ =\frac{Q_{\mathrm{in}}}{{\mathrm{Ï\mu }}_o} \end{gathered}$$

the electrical flux depends on the charge inside the closed surface. any flux due to charges outside the closed surface is zero

Pertaining to the enclosed surface,

${\mathrm{Φ}}_{\mathrm{A}}=\frac{-q}{{\mathrm{Ï\mu }}_o}$, As a result, the surface's net charge is

$q_1+q_3=-q$ $\left(1\right)$

pertaining to the enclosed surface,

${\mathrm{Φ}}_{\mathrm{B}}=\frac{3q}{{\mathrm{Ï\mu }}_o}$, As a result, the surface's net charge is

$q_1+q_2=3q$ $\left(2\right)$

pertaining to the enclosed surface,

${\mathrm{Φ}}_C=\frac{-2q}{{\mathrm{Ï\mu }}_o}$, As a result, the surface's net charge is

$q_3+q_2=-2q$ $\left(3\right)$

Rewrite the equation $\left(2\right)$ in a different way

$q_1=3q-q_2$

Use this in equation $\left(1\right)$

$q_1+q_3=-q$

$3q-q_2+q_3=-q$

$q_3-q_2=-4q$ $\left(4\right)$

Add equation $\left(3\right)$and $\left(4\right)$to get $q_3$

$q_3+q_2=-2q$

$+$

$q_3-q_2=-4q$

$2q_3=-6q$

$q_3=-3q$

Put values of $q_3$in equation $\left(1\right)$

$$\begin{gathered} q_1=-q-q_3 \\ =-q-(-3q) \\ =2q \end{gathered}$$

Use values of $q_1$in equation $\left(2\right)$

$$\begin{gathered} q_2=3q-q_1 \\ =3q-2q \\ =q \end{gathered}$$