JEE Main 2026 Apr 2 Shift 1, Physics Q1: Units And Dimensions

JEE Main2026Apr 2, Shift 1Physics

The dimensional formula of $\frac{1}{2} ε_{0} E^{2}$ ( $ε_{0}$ = permittivity of vacuum and $E$ = electric field) is $M^{a} L^{b} T^{c}$ . The value of $2 a - b + c =$ ________.

A
$0$
B
$1$
C
$- 1$
D
$2$

Solution

The quantity $\frac{1}{2} ε_{0} E^{2}$ represents the electric field energy density (energy per unit volume).

Therefore its dimensions are those of energy divided by volume:

$[\frac{1}{2} ε_{0} E^{2}] = \frac{[ M L ^{2} T ^{- 2} ]}{[ L ^{3} ]} = M L^{- 1} T^{- 2}$

Comparing with $M^{a} L^{b} T^{c}$ , we get $a = 1$ , $b = - 1$ , $c = - 2$ .

$∴ 2 a - b + c = 2 (1) - (- 1) + (- 2) = 2 + 1 - 2 = 1$

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JEE Main 2026 Apr 2 Shift 1, Physics Q1: Units And Dimensions

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