NTA Abhyas JEE Main2020PhysicsUnits and DimensionsPractice
Find the dimensions of electric permittivity
Options
- AA 2 M - 1 L - 3 T 4
- BA 2 M - 1 L - 3 T 0
- CA M - 1 L - 3 T 4
- DA 2 M 0 L - 3 T 4
Correct answer
A. A 2 M - 1 L - 3 T 4
Step-by-step solution
From Coulomb's law, the force of attraction/repulsion between two point charges q and q separated by distance r is                                           F = 1 4 π ε 0 q 2 r 2 ⇒                               ε 0 = 1 4 π   . q 2 F r 2 Where ε 0 is electric permittivity. Dimensions of ε 0 = A