The dimensional formula of 12ε0E2 (ε0: permittivity of free space and E: Intensity of electric field) is

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The dimensional formula of $\frac{1}{2} ε_{0} E^{2}$ ( $ε_{0}$ : permittivity of free space and $E :$ Intensity of electric field) is

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$[M L^{2} T^{- 2}]$

$[M L^{2} T^{- 1}]$

$[M L T^{- 1}]$

$[M L^{- 1} T^{- 2}]$

answer is C.

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Detailed Solution

Given
Energy density of electric field $= \frac{1}{2} ε_{0} E^{2}$
Here $ε_{0} =$ Permittivity of free space
$E =$ Intensity of electric field
$\Rightarrow [\frac{1}{2} ε_{0} E^{2}] = [\frac{Energy}{Volume}] = \frac{M L^{2} T^{- 2}}{L^{3}} = M L^{- 1} T^{- 2}$
$∴ Dimension formula of \frac{1}{2} ε_{0} E^{2} = M L^{- 1} T^{- 2}$
Therefore, the correct answer is (C).