Using mass `(M)` , length `(L)` , time `(T)` , and electric current `(A)` as fundamental quantities , the dimensions of permitivity will be

Options

A. `[M LT^(-1) A^(-1)]`
B. `[M LT^(-2) A^(-1)]`
C. `[M^(-1) L^(-3) T^(4) A^(2)]`
D. `[M^(2) L^(-2)T^(-2) A]`

Answer

C

Explanation

By Coulomb's law , `F = (1)/( 4 pi epsilon_(0)) xx ( q_(1) q_(2))/( r^(2))`
`epsilon_(0)=(q_(1)q_(2))/(4pixxFxxr^(2))`
Taking dimensions, `epsilon_(0) = (( A T) ( A T)) /( ML^(1) T^(-2) xx L^(2)) = [ M^(-1) L^(-3) T^(4) A^(2)]`