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JEE Main 8 April 2024, Shift 1, Physics Q55

Question 55 of 90 in this shift, Physics question 25 of 30, Section B.

An electric field,

\vec{E}=\frac{2\hat{i}+6\hat{j}+8\hat{k}}{\sqrt{6}}

passes through the surface of 4 m

^2

area having unit vector

\hat{n}=\left(\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)

. The electric flux for that surface is ________ V m.

Official answer

NTA final key.

Chapter: Electric Charges and Fields
Topic: Gauss's Law
Idea tested: Electric Flux Calculation

Same idea in other shifts

Asked 3× in all

7 Apr 2025, Shift 2 · Q50The electric field in a region is given by $\vec{E}=\left(2\hat{i}+4\hat{j}+6\hat{k}\right)\times10^3$ N/C. The flux of the field through…EasyNumerical value
9 Jan 2020, Shift 2 · Q22An electric field $\vec{E} = 4x\,\hat{i} - (y^2+1)\,\hat{j}$ N/C passes through the box shown in figure. The flux of the electric field…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.