Physics

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



      


ANSWER

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


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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