what is the electric force between +2.5uC and -5uC separated by a distance of 5cm?

Answer 1

It is given by Coulomb's Law.
Formula is:-
F=(1/4*pi*epsilon)*(q1*q2/r^2)
Where epsilon:-permittiviy of medium
In vacuum epsilon=8.85*10^-12
1/4*pi*epsilon=9*10^9

GIven:-
r=5cm=5*10^-2m
q1=2.5uC=2.5*10^-6
q2=-5*10-6

All to do is calculations.
As both charges are of opposite signs the force will be attractive.

Answer 2

Hi Good Tariq M

Answer is = - 45 Newton (attractive force)

Give marks to Tariq