If εo is the permittivity of free space and E is the electric field, then εoE² has the dimensions: (a) [ML⁻¹T⁻²] (b) [ML²T⁻²] (c) [M⁰L⁻²TA] (d) [M⁻¹L⁻³T⁴A²]

Answer: (a). 1/2ε₀E² is the energy density of an electric field (energy per unit volume). [ε₀E²] = ML²T⁻²/L³ = ML⁻¹T⁻² So the answer is (a) [ML⁻¹T⁻²].

If εo is the permittivity of free space and E is the electric field… | JEE Main 2024 (8 Apr, Shift 2)

Q26JEE Main 2024 · 8 APR · SHIFT 2Easy

Units & Dimensions

If $\epsilon_o$ is the permittivity of free space and $\mathrm{E}$ is the electric field, then $\epsilon_o\mathrm{E}^2$ has the dimensions:

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Answer: (a)

$\frac{1}{2}\epsilon_0E^2$ is the energy density of an electric field (energy per unit volume).
$[\epsilon_0E^2] = \mathrm{ML^2T^{-2}}/\mathrm{L^3} = \mathrm{ML^{-1}T^{-2}}$
So the answer is (a) $[\mathrm{ML^{-1}T^{-2}}]$ .

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