Question

Following figure shows for Gaussian surface $S_{1}, S_{2}, S_{3}$ and $S_{4}$.

Now, match the following columns and mark the correct code given below.

Column I		Column II
i	$\phi_{S_{1}}$	p	$+\frac{Q}{\varepsilon_{0}}$
ii	$\phi_{S_{2}}$	q	0
iii	$\phi_{S_{3}}$	r	$-\frac{Q}{\varepsilon_{0}}$
iv	$\phi_{S_{4}}$	s	$-\frac{2Q}{\varepsilon_{0}}$

• A. i- (s) , ii - (q) , iii - (r) , iv - (p)
• B. i - (q), ii - (s), iii - (r), iv - (p)
• C. i - (q), ii - (r), iii - (s), iv - (p)
• D. i - (r) , ii - (s) , iii - (p) , iv - (q)

Answer 1

Answer: (A)

The net flux through each closed surface is determined by the net charge inside. $\phi_{s_{1}}=\frac{(+Q-3 Q)}{\varepsilon_{0}}=\frac{-2 Q}{\varepsilon_{0}}$
$\phi_{s_{2}}=\frac{(+Q+2 Q-3 Q)}{\varepsilon_{0}}=0$
$\phi_{s_{3}}=\frac{(+2 Q-3 Q)}{\varepsilon_{0}}=-\frac{Q}{\varepsilon_{0}}$
$\phi_{s_{4}}=+\frac{Q}{\varepsilon_{0}}$