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JEE Main 2026 · 2 Apr · Shift 1, Physics question 26

The dimensional formula of 12ϵ0E2\dfrac12\epsilon_0E^2 (ϵ0\epsilon_0 = permittivity of vacuum and EE = electric field) is MaLbTcM^aL^bT^c. The value of 2a−b+c2a-b+c = ________.

Choose one option
Answer and worked solution

Official answer: B

Energy density. 12ϵ0E2\frac12\epsilon_0E^2 is energy per unit volume: ML2T−2L3=ML−1T−2\dfrac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}.

Exponents. a=1a=1, b=−1b=-1, c=−2c=-2, so 2a−b+c=2+1−2=12a-b+c=2+1-2=1.

(B)

Topics:Dimensional analysisEnergy stored in a capacitor
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