Question

Three point charges lie in a straight line along the y-axis. a charge of q1 = -9.10 µc is at y = 6.30 m, and a charge of q2 = -7.90 µc is at y = -4.50 m. the net electric force on the third point charge is zero. where is this charge located?

Answer 1

Electric field E = kQ/r^2

Distance between charges = 6.30 - (-4.40) = 10.70m

Say the neutral point, P, is a distance d from q1. This means it is a distance (10.70 - d) from q2.

Field from q1 at P = k(-9.50x^10^-6) / d^2

Field from q2 at P = k(-8.40x^10^-6) / (10.70-d)^2

These fields are in opposite directions and are equal magnitudes if the resultant field = 0

k(-9.50x^10^-6) / d^2 = k(-8.40x^10^-6) / (10.70-d)^2

9.50 / d^2 =8.40 / (10.70-d)^2

d^2 / (10.70-d)^2 = 9.50/8.40 = 1.131

d/(10.70-d) = sqrt(1.1331) = 1.063

d = 1.063 ((10.70-d)

= 10.63 - 1.063d

2.063d = 10.63

d = 5.15m

The y coordinate where field is zero is 6.30 - 5.15 = 1.15m

Step-by-step explanation: