Question

particle 1 carries 4.50 uC of electric charge. Particle 2 carries 2.63 uC of electric charge. Particles 1 & 2 are 11.8 cm apart. Find the net force of a particle 3 carrying 3.16 uC of charge and located 3.54 cm from Particle 1. in Newtons.

Answer 1

Given

charges are q1 = 4.5*10^-6 C ,

q2 = 2.63*10^-6 C ,

q3 = 3.16*10^-6 C ,

q1,q2 are separated by a distance of r12 = 11.8 cm

q1,q3 are separated by 3.54 cm = 0.0354 m

what is the force on q1 by q3

ler the charges are q3 ----- q1 ----------------------q2

From Coulomb's law F = kq1*q2/r^2

F13 = kq1*q3/r13^2

and F23 = kq2*q3/r23^2

F13 = (9*10^9)(4.5*10^-6)(3.16*10^-6)/(0.0354)^2 N = 102 N

F23 = (9*10^9)(2.63*10^-6)(3.16*10^-6)/(0.0354+0.118)^2 N = 3.178 N

the net force is F = F13+F23 = 102+3.178 N = 105.178 N