At what coordinates can a third charge be placed so that it experiences no net force?

Question

Here's the entire question:
A +5.2 µC charge is placed at the origin and a -2.7 µC charge is placed at x = 25 cm. At what coordinates can a third charge be placed so that it experiences no net force?

Here's the hint my teacher gave us:
The point HAS to be along a line connecting the two charges. But it doesn't have to be BETWEEN them. Do this one like we did the force = zero on a third mass. But instead of x and r-x for distances, you use x and r+x if it is OUTSIDE.

I have been trying to do this problem for ages, but still can't get the right answer. Can someone help please? Thank you for your time.

Answer 1

actually the problem is the data for calculation

u can du it like

assume the third charge as q

k*q*5.2/(25+x)^2 = k* q* 2.7/x^2

after solving it i got answer as

x=6*(39)^1/2+27

Answer 2

The teacher has practically told you how to do it. Put the 3rd charge on the OTHER side of charge that's at x = 25 cm. So, as the teacher said, use distances r for the weaker charge, and (25 + r) for the stronger one, and because the charges are opposite, make them balance out. To wit:

(2.7)^2 / r^2 = (5.2)^2 / (25 + r)^2

Solve for r. When the messy math is done, you get r = 27, so the magic spot is at x = 52. To double check:

(2.7)^2 / (27)^2 = 0.01 = (5.2)^2 / (52)^2

Answer 3

The point is obviously on a straight line connecting all 3 charges, but it is to the right of the -2.7uC charge, as you want the weaker of the source charges to be closer.

As to the exact distance, that is simple to calculate, since force varies with the square of the distance, (25+x)^2 * 5.2 = (x)^2 * 2.7, where x is the distance to the right of the 2.7 uC charge, or something similar, you figure out the rest for the homework.