# Homework Help: Coulomb's Law and distance

1. Jan 9, 2010

### Larrytsai

1. The problem statement, all variables and given/known data
Two charges, one of 2.50 micro coulomb, and the other of -3.50 micro coulomb, are placed on the x-axis, one at the origin and the other at x=0.600m, as shown in Fig 21.36. Find the position on the x-axis where the net force on a small charge +q would be zero

2. Relevant equations
F= kq1q1/r^2

3. The attempt at a solution
I tried to equate the force when the distance of the two charges are 0.6m away to the force when the distance is x+0.6

2. Jan 9, 2010

### ideasrule

You can imagine that the charge has to be on the negative x axis for the force exerted by the positive charge to equal the force exerted by the negative charge. Assume that it's a distance "d" from the origin; then it would be d+0.6 m from the other charge. Proceed from there.

3. Jan 9, 2010

### Larrytsai

yeah i have made that assumption by equating F1=F2, but, i dont understand whats happening after, i cancel my constants, and my charges should cancel so i end up with 1/(0.6)^2 = 1/(x+0.6)^2 , and u end up with 0

4. Jan 9, 2010

### ideasrule

How do the charges cancel? One's 3.5 mC (in magnitude) and the other's 2.5 mC.

5. Jan 9, 2010

### Larrytsai

well i have F1=F2 so its kq1q2/(r)^2 = kq1q2/(r+6)^2

6. Jan 10, 2010

### ideasrule

Don't apply equations without any thought. Think about what the letters in this equation:

kq1q2/(r)^2 = kq1q2/(r+6)^2

stand for. Assuming q1 is the test charge, do the q2's represent the same charge?

7. Jan 10, 2010

### Larrytsai

im sorry but i do not understand the last part, if we assume q1 is the test charge, so we are able to cancel q1 out?