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JEE Main 2021 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 3 August 2021, Shift 2, Physics Q29

Question 29 of 90 in this shift, Physics question 29 of 30, Section B.

The electric potential at point

(x, 0, 0)

is

V = \left(\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3}\right)

volt. The electric field strength at a point

x = 1\ m

is

\underline{\hspace{2em}}\ \hat{i}\ V/m

.

Official answer

5500

NTA final key.

Chapter: Electrostatic Potential and Capacitance
Topic: Electric Potential
Idea tested: Field-Potential Relation

Same idea in other shifts

Asked 2× in all

6 Apr 2026, Shift 2 · Q37The electric potential as a function of $x, y$ is given by $V=5(x^2-y^2)$ V. The electric field at a point (2, 3) m is __________ V/m.EasySingle correct

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.