Question

If a -1 nC point charge is placed 2.0 m away from a 1 nC point charge, what is the magnitude and direction of the force acting on the negative charge?

Answer 1

Given that q1=-1nC

q2=1nC

R=2.0m

Our equation is from Coulomb's law

F=(k* modulous of q1* modulous of q2)/R^2

where k=(1/4*pie*epsilon zero)=constant =8.99*10^9N. m^2/C^2

Substituting we get

Magnitude of F=(8.99*10^9*1*10^9*1*10^9)/2.0)^2 Note that here we take modulous value.

We can write magnitude of force

F=2.2475nN

Because two of the charge are in opposite direction, they attract each other

So force is directed towards q1

Therefore direction of the force

F =+ 2.2475nN i^ in unit vector notation