Question

Two point charges, –8.0 nC and +16 nC , are held fixed a distance 0.60 m apart. Assume that both charges lie on x-axis, for the -8.0 nC x = 0 and for the +16 nC charge x = 0.60 m . Where can a +20 nC charge be placed so that the net force on it is zero? Repeat part A if the third charge is –20 nC.

Answer 1

q1 = -8 nC
q2 = 16 nC
q3 = 20 nC
kq1q3/x^2 = kq2q3/(0.6+x)^2
8*20/x^2 = 16*20/(0.6+x)^2

(0.6+x)^2 = 2x^2
x = -0.2485 m

q1 = -8 nC
q2 = 16 nC
q3 = -20 nC
kq1q3/x^2 = kq2q3/(0.6+x)^2
8*20/x^2 = 16*20/(0.6+x)^2

(0.6+x)^2 = 2x^2
x = -0.2485 m