Question

Consider three point charges \(-2Q, Q\) and \(-Q\) and three surfaces S\(_1\), S\(_2\) and S\(_3\) as shown in the figure. Match the entries of List-I with that of List-II.

List-I
(a) Net flux through S\(_1\)
(b) Net flux through S\(_2\)
(c) Net flux through S\(_3\)
List-II
(i) \(\frac{-2Q}{\epsilon_0}\)
(ii) \(\frac{-Q}{\epsilon_0}\)
(iii) Zero
Codes:

• A. a - ii, b - i, c - iii
• B. a - iii, b - ii, c - i
• C. a - i, b - ii, c - iii
• D. a - ii, b - iii, c - i

Hint:

Charges located outside a Gaussian surface do not contribute to the net flux through that surface, regardless of their magnitude or proximity.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem requires calculating the net electric flux through different Gaussian surfaces by identifying which charges are enclosed within each surface.
Step 2: Key Formula or Approach:
By Gauss's Law, \( \Phi = \frac{\sum q_{in}}{\epsilon_0} \), where \( \sum q_{in} \) is the algebraic sum of charges inside the closed surface.
Step 3: Detailed Explanation:
From the figure:
1. Surface S\(_1\) encloses the charges \(-2Q\) and \(Q\).
Net flux through S\(_1\) = \(\frac{-2Q + Q}{\epsilon_0} = \frac{-Q}{\epsilon_0}\).
This matches with (ii).
2. Surface S\(_2\) encloses the charges \(Q\) and \(-Q\).
Net flux through S\(_2\) = \(\frac{Q - Q}{\epsilon_0} = 0\).
This matches with (iii).
3. Surface S\(_3\) is the outermost surface and encloses all three charges: \(-2Q, Q\), and \(-Q\).
Net flux through S\(_3\) = \(\frac{-2Q + Q - Q}{\epsilon_0} = \frac{-2Q}{\epsilon_0}\).
This matches with (i).
Therefore, the correct matching is: a - ii, b - iii, c - i.
Step 4: Final Answer:
The correct code is (4).