Physical World Units and Measurements - Result Question 14
14. The dimension of $\frac{1}{2} \varepsilon_0 E^{2}$, where $\varepsilon_0$ is permittivity of free space and $E$ is electric field, is:
(a) $[ML^{2} T^{-2}]$
(b) $[ML^{-1} T^{-2}]$
(c) $[ML^{2} T^{-1}]$
(d) $[MLT^{-1}]$
[2010]
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Answer:
Correct Answer: 14. (b)
Solution:
- (b) $\frac{1}{2} \varepsilon_0 E^{2}$ represents energy density i.e., energy per unit volume. $\Rightarrow[\frac{1}{2} \varepsilon_0 E^{2}]=\frac{[M L^{2} T^{-2}]}{[L^{3}]}=[ML^{-1} T^{-2}]$
