JEE Main2013PhysicsUnits and DimensionsActual
Let ∈ 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then :
Options
- A∈ 0 = M -1 L 2 T -1 A - 2
- B∈ 0 = M -1 L 2 T -1 A
- C∈ 0 = M -1 L - 3 T 2 A
- D∈ 0 = M -1 L - 3 T 4 A 2
Correct answer
D. ∈ 0 = M -1 L - 3 T 4 A 2
Step-by-step solution
F = 1 4 π ε 0 q 1 q 2 R 2 ε 0 = q 1 q 2 4 π FR 2 Hence, ε 0 = C 2 N . m 2 = AT 2 ML T - 2 . L 2 = M -1 L - 3 T 4 A 2