Question

Using mass(m),length(l),time(t) and electric current(a) as fundamental quantities,the dimensions of permittivity will be

Answer 1

Answer:

Explanation:

Let us consider Coulomb's law : F = q2 /(4πε0×r2)  ;  

q is charge in Coulomb, r is distance in metre, F is force in Newton  and ε0 is permeability.

 

dimenison of ε0 = dimension of q2/(F×r2) = [Coulomb]2 /[Newton× square metre]................(1)

 

dimension of Coulomb = dimension of (Current×Time) = [ A T ]

dimension of Newton  = [ MLT-2]

 

substitute the above in (1), we have

 

dimension of ε0 = [ A2T2 ]/ [MLT-2 × L2] = [ A2 M-1 L-3 T4 ]

Answer 2

Answer:

M^-1L^-3A^2T^4

Explanation:

We know,

Force=F=q^2/4π€r^2

Then, €=q^2/4πFr^2

€=AT×AT/MLT^-2*L^2

€=M^-1L^-3A^2T^4