the electric force experienced by a charge of 1*10^-6 c is 1.5*10^-3 N. find the magnitude of the electric field at the position of the charge?

Charge(q)=1*10^-6c

Force(F)=1.5*10^-3N

electric field(E)=F/q=(1.5×10^−3)÷(1×10^−6) =1500N/C

I think you know the relation between  force,charge and electric field.Electric field is defined as force per unit charge.

i.e F=E*q

E=F/q

 =(1.5×10^(−3))/(1×10^(−6))

 =1.5*10^3

 =1500N/C

This is the formula for this question

K= 9*10^9

The Answer is 2.449489743