Question

Three charges of $q_1=1\times {10}^{-6}\text{C}$, $q_2=2\times {10}^{-6}\text{C}$ and $q_3=-3\times {10}^{-6}\text{C}$ have been placed as shown. Then, the net electric flux will be maximum for the surface

• A. ${\text{S}}_1$
• B. ${\text{S}}_2$
• C. ${\text{S}}_3$
• D. Same for all three

Answer 1

The correct option is A ${\text{S}}_1$

According to the Gauss's law the flux of the electric field through closed surface is
$\varphi =\frac{q_{\text{en}}}{{\epsilon }_0}$
Where, $q_{en}$ is the charge enclosed by the closed surface.

So, the flux will be maximum for that surface, which incloses maximum charge.

Charge enclosed by surface ${\text{S}}_1$,
$Q_{{\text{S}}_1}=q_1+q_2$
$\Rightarrow Q_{{\text{S}}_1}=1\times {10}^{-6}+2\times {10}^{-6}$
$\Rightarrow Q_{{\text{S}}_1}=3\times {10}^{-6}\text{C}$

Charge enclosed by surface ${\text{S}}_2$,
$Q_{{\text{S}}_2}=q_2+q_3$
$\Rightarrow Q_{{\text{S}}_2}=2\times {10}^{-6}-3\times {10}^{-6}$
$\Rightarrow Q_{{\text{S}}_2}=-1\times {10}^{-6}\text{C}$

Charge enclosed by surface ${\text{S}}_3$,
$Q_{{\text{S}}_3}=q_1+q_2+q_3$
$\Rightarrow Q_{{\text{S}}_3}=1\times {10}^{-6}+2\times {10}^{-6}-3\times {10}^{-6}$
$\Rightarrow Q_{{\text{S}}_3}=0$

So, the charge enclosed is maximum across ${\text{S}}_1$
So, flux will be maximum across ${\text{S}}_1$.

Hence, option (a) is correct asnwer.