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Saturday, 31 October

JEE Main · Physics · Electrostatic Potential and Capacitance

Field-Potential Relation

JEE Main asked Field-Potential Relation 3 times in 3 of the 108 shifts Lumi analysed, in 2021 and 2026. It belongs to Electric Potential, in Electrostatic Potential and Capacitance (Physics). The longest wait between two years asking it was 5 years. In 2021 it came in more than one shift of the same year.

Updated 2026-10-24

Times asked
3
in 3 shifts
Years
2 of 8
First asked
2021
last in 2026
Pattern
Occasional
across the analysed shifts

Questions on this idea a shift, by year

2017–2026. Asked in 2021 and 2026. Years differ in how many shifts Lumi analysed, so each year is scaled to one shift of 25 Physics questions.

00.10.20’170’200.07’210’220’230’240’250.11’26

Raw count: 2017: 0 in 1 shift · 2020: 0 in 12 shifts · 2021: 2 in 24 shifts · 2022: 0 in 24 shifts · 2023: 0 in 12 shifts · 2024: 0 in 18 shifts · 2025: 0 in 8 shifts · 2026: 1 in 9 shifts.

Shift by shift

Each square is one analysed shift, in date order. Open a filled one for that paper.

  • 20261 of 9
  • 20250 of 8
  • 20240 of 18
  • 20230 of 12
  • 20220 of 24
  • 20212 of 24
  • 20200 of 12
  • 20170 of 1

Asked in that shift More than once Not asked

How JEE Main framed it

Single correct is the usual form. Skill tested: calculation.

  • Field vector from potential function
  • Field from potential function

Difficulty

Easy 3Medium 0Hard 0

1 of the 3 questions had a numerical answer.

Every question

All 3 questions

From the official JEE Main shift papers, newest first. Open one for the full question, the options and the official answer.

  1. 6 Apr 2026, Shift 2 · Q37The electric potential as a function of x,yx, y is given by V=5(x2−y2)V=5(x^2-y^2) V. The electric field at a point (2, 3) m is __________ V/m.EasySingle correct
  2. 4 Aug 2021, Shift 2 · Q9The electric potential at any point P(x, y, z) is V=x3z−x2y−3V=x^3z-x^2y-3 volts. The electric field E⃗Q\vec{E}_Q at point Q(2, 3, 1) will be (in…EasySingle correct
  3. 3 Aug 2021, Shift 2 · Q29The electric potential at point (x,0,0)(x, 0, 0) is V=(1000x+1500x2+500x3)V = \left(\frac{1000}{x}+\frac{1500}{x^2}+\frac{500}{x^3}\right) volt. The electric field…EasyNumerical value

Shifts: 3 Aug 2021, Shift 2, 4 Aug 2021, Shift 2, 6 Apr 2026, Shift 2.

Same topic

Other ideas in Electric Potential

Electric Potential has 16 ideas JEE Main has asked; 7 of them came up in more than one question. The 12 most asked are below.

Topic analysisChapter analysisStudy this chapterAll repeated ideas