JEE Main 2025 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 7 April 2025, Shift 2, Physics Q50

Question 50 of 75 in this shift, Physics question 25 of 25, Section B.

The 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 a rectangular surface parallel to

x

z

plane is

6.0\ \mathrm{Nm^2C^{-1}}

. The area of the surface is __________ cm

^2

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

8 Apr 2024, Shift 1 · Q55An 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…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.