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Friday, 9 October

JEE Main 2020 · PhysicsMultiple choiceSingle correctMediumConceptual

JEE Main 8 January 2020, Shift 1, Physics Q13

Question 13 of 75 in this shift, Physics question 13 of 25, Section A.

In finding the electric field using Gauss law the formula ∣E⃗∣=qencϵ0∣A∣\left|\vec{E}\right| = \frac{q_{enc}}{\epsilon_0 |A|} is applicable. In the formula ϵ0\epsilon_0 is permittivity of free space, A is the area of Gaussian surface and qencq_{enc} is charge enclosed by the Gaussian surface. This equation can be used in which of the following situation ?
  1. (1)For any choice of Gaussian surface.
  2. (2)Only when the Gaussian surface is an equipotential surface.
  3. (3)Only when the Gaussian surface is an equipotential surface and ∣E⃗∣\left|\vec{E}\right| is constant on the surface.Official answer
  4. (4)Only when ∣E⃗∣\left|\vec{E}\right| = constant on the surface.

Official answer

Option 3

NTA final key (Jan 2020).

Same topic in other shifts

All Gauss's Law questions
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  3. 8 Apr 2025, Shift 2 · Q36An infinitely long wire has uniform linear charge density λ=2\lambda = 2 nC/m. The net flux through a Gaussian cube of side length…MediumSingle correct
  4. 8 Apr 2024, Shift 1 · Q55An electric field, E⃗=2i^+6j^+8k^6\vec{E}=\frac{2\hat{i}+6\hat{j}+8\hat{k}}{\sqrt{6}} passes through the surface of 4 m2^2 area having unit vector…EasyNumerical value
  5. 9 Apr 2024, Shift 2 · Q42Five charges +q+q, +5q+5q, −2q-2q, +3q+3q and −4q-4q are situated as shown in the figure. The electric flux due to this configuration through…EasySingle correctHas a figure
  6. 9 Apr 2024, Shift 2 · Q56An electric field E⃗=(2xi^) NC−1\vec{E}=(2x\hat{i})\ NC^{-1} exists in space. A cube of side 22m is placed in the space as per figure given below.…MediumNumerical valueHas a figure

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.