Calculating net force on a charge placed at a spot

[phys62]

Homework Statement

Two charges, -16 and +3.4 mC, are fixed in place and separated by 4.4 m. (a) At what spot along a line through the charges is the net electric field zero? Give the distance of the spot to the positive charge in meters (m). (Hint: The spot does not necessarily lie between the two charges.) (b) What would be the force on a charge of +14 mC placed at this spot?

Homework Equations

E=kq/r^2

The Attempt at a Solution

I've figured out part a, but I don't know what I'm doing wrong on part b... here is my work:

a) the correct answer is 3.76 m.

b) F21=(8.99x10^9)(3.4x10-6)(14x10-6)/(3.76 m)^2 = 0.03026 N
F23=(8.99x10^9)(-16x10^-6)(14x10-6)/(4.4+3.76)^2 = -0.0302 N

then I added the 2 together and got 1.68x10^-5 N
Am I supposed to actually get 0 as my answer? I'm so confused.

 

[chrisk]

F=qE. Because E = 0 at this spot any charge placed at this spot will not experience a force. It's analogous to zero gravity; any mass placed in a zero gravity field will not experience a force.

 

[thomate1]

I can confirm that your calculations for part a are correct. The net electric field at a distance of 3.76 m from the positive charge is zero, as the repulsive force from the positive charge is balanced by the attractive force from the negative charge.

For part b, you are correct in using the equation F=ke(q1q2)/r^2 to calculate the force between the two charges. However, you should first calculate the distance between the charge of +14 mC and the two fixed charges (-16 mC and +3.4 mC) at the spot where the net electric field is zero. This distance will be different from the 3.76 m you calculated in part a.

Once you have the correct distance, you can plug it into the equation to calculate the force on the +14 mC charge. The resulting force should be zero, as the net electric field at this spot is zero and there is no force acting on the charge.