Electrostatics Question 426

Question: The dimension of (1/2) $ {\varepsilon _{0}}E^{2}({\varepsilon _{0}} $ : permittivity of free space; $ E $ : electric field) is [IIT-JEE (Screening) 2000; KCET 2000]

Options:

A) $ ML{{T}^{^{-1}}} $

B) $ ML^{2}{{T}^{-2}} $

C) $ M{{L}^{-1}}{{T}^{-2}} $

D) $ ML^{2}{{T}^{-1}} $

Show Answer

Answer:

Correct Answer: C

Solution:

Energy density $ =\frac{\text{Energy}}{\text{Volume}} $ so it’s dimensions are $ \frac{ML^{2}{{T}^{-2}}}{L^{3}}=[M{{L}^{-1}}{{T}^{-2}}] $



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