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

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Units & Measurement

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 =

____.

Answer: (B)

The quantity

\frac{1}{2} ϵ_{0} E^{2}

is the energy density of the electric field, i.e. energy per unit volume, with dimensions

\frac{[ M L ^{2} T ^{- 2} ]}{[ L ^{3} ]} = M L^{- 1} T^{- 2}

. Thus

a = 1

b = - 1

c = - 2

, and

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

. The value of

2 a - b + c

1

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option