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}}] $
