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  4. The dimension of (1/2)ε 0 E2, where ε 0 is permittivity of free space and E is electric field, is

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Q. The dimension of $\frac{1}{2}\varepsilon _0 E^2$, where $\varepsilon _0$ is permittivity of free space and $E$ is electric field, is

AIPMTAIPMT 2010Physical World, Units and Measurements

Solution:

Energy density of an electric field $E$ is
$u_E=\frac{1}{2}\varepsilon _0 E^2$
where $\varepsilon _0$ is permittivity of free space
$u_E=\frac{\text{Energy}}{\text{Volume}}$
$=\frac{ML^2T^{-2}}{L^3}$
$=ML^{-1}T^{-2}$
Hence, the dimension of $\frac{1}{2}\varepsilon _0 E^2 $ is $ML^{-1}T^{-2}$