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JEE Main 2026 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 2 April 2026, Shift 1, Physics Q26

Question 26 of 75 in this shift, Physics question 1 of 25, Section A.

The dimensional formula of

\frac12\epsilon_0E^2

(

\epsilon_0

= permittivity of vacuum and

E

= electric field) is

\mathrm{M^aL^bT^c}

. The value of

2a-b+c=

________.

(1) $0$
(2) $1$ Official answer
(3) $-1$
(4) $2$

Official answer

Option 2

NTA final key.

Chapter: Units and Measurements
Topic: No close match in our taxonomy
Idea tested: No close match in our taxonomy

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.