Question

Four closed surface and corresponding charges distributions are shown below:
Let the respective electric fluxes through the surfaces be ${\Phi }_1$, ${\Phi }_2$, ${\Phi }_3$ and ${\Phi }_4$. Then :

• A. ${\Phi }_1>{\Phi }_3;{\Phi }_2<{\Phi }_4$
• B. ${\Phi }_1<{\Phi }_2={\Phi }_3>{\Phi }_4$
• C. ${\Phi }_1>{\Phi }_2>{\Phi }_3>{\Phi }_4$
• D. ${\Phi }_1={\Phi }_2={\Phi }_3={\Phi }_4$

Answer 1

Answer: (D)

Flux through a surface is given by $\varphi =\frac{\Sigma q}{ϵ}$, where $\Sigma q=$ Total charge enclosed by surface

$\Rightarrow {\varphi }_1=\frac{2q}{ϵ}$

$\Rightarrow {\varphi }_2=\frac{2q}{ϵ}$

$\Rightarrow {\varphi }_3=\frac{2q}{ϵ}$

$\Rightarrow {\varphi }_4=\frac{2q}{ϵ}$

$\Rightarrow {\varphi }_1={\varphi }_2={\varphi }_3={\varphi }_4$

Reason: Outside charges have no effect on the surface (Gauss law)

So, option D is correct.