Physics

The dimension of $\cfrac {1}{2}\epsilon_o E^2$ permittivity of free space and E:Intensity of electric field) is

$[ML^{-1}T^{-2}]$

SOLUTION
Energy density of an electric field $E$ is ,

$\mu_E=\dfrac 12 \epsilon_0 E^2$

where $\epsilon_0$ is permitivity of free space

$\mu_E=\dfrac{Energy}{Volume}=\dfrac{[ML^2T^{-2}]}{[L^3]}=[ML^{-1}T^{-2}]$

Hence, the dimension of $\dfrac 12 \epsilon_0 E^2$ is $[ML^{-1}T^{-2}]$

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Single Correct Medium Published on 18th 08, 2020
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