Matching Charge Distributions with Flux and Field – Rankers Physics

Gauss's Law: Practice Problem & Solution

Column A contains some charge distribution and column B contains corresponding electric flux or field. Match the columns and choose the correct option. Column A: A. Charge outside a closed gaussian surface B. Charge \(q\) inside a closed gaussian surface C. Infinite plane sheet of charge D. Field outside a charged conducting sphere Column B: (P) \(\oint \vec{E} \cdot d\vec{A} = \frac{q}{epsilon_0}\) (Q) \(E = \frac{\sigma}{2\epsilon_0}\) (R) \(E = \frac{KQ}{r^2}\) (S) Net flux is zero  
A(S), B(P), C(R), D(Q)
A(S), B(P), C(Q), D(R)
A(P), B(R), C(Q), D(S)
A(P), B(Q), C(S), D(R)

Solution Explained:

To solve this problem, we apply the core principles of Gauss's Law. Understanding the underlying formula is key to arriving at the correct answer below:

A charge outside a closed surface contributes zero net flux (A-S). Gauss's law states that for an enclosed charge \(q\), the net flux is \(\frac{q}{\epsilon_0}\) (B-P). The electric field due to an infinite plane sheet is \(E = \frac{\sigma}{2\epsilon_0}\) (C-Q). For a charged conducting sphere, the external field is \(E = \frac{KQ}{r^2}\) (D-R).

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