Let ε 0 denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then dimensions of permittivity is given as M p L q T r A s . Find the value of p - q + r s

Step-by-step solution

From Coulomb's law, F = q 1 q 2 4 π ε 0 r 2 ∴ ε 0 = q 1 q 2 4 π F r 2 = ( A 1 T 1 ) ( A 1 T 1 ) [ M 1 L 1 T - 2 ] [ L 2 ] = M - 1 L - 3 T 4 A 2 ∴ p = - 1 , q = - 3 , r = 4 , s = 2 ∴ p - q + r s = - 1 - ( - 3 ) + 4 2 = 6 2 = 3