Question

Consider two charges, where one (+1.90 nC) is at the origin and the other (-3.61 nC) is at the position x = 4.35 cm.

Find the x-coordinate where a proton would experience zero net force.

Answer 1

here.... +small ...... -big

x ......... 0.......... 4.35

just using distances without regard to sign to get equal field magnitudes
1.90/x^2 = 3.61/(x+4.35)^2

solve for x, note it has to be - (left of origin)

3.61 x^2 = 1.9(x^2 +8.7 x+18.9)
3.61x^2 = 1.9x^2 + 16.5x + 35.9
1.71x^2 -16.5x -35.9 = 0

|x| = either 11.4 or 1.83
try -1.83
check
.567 = 1.9/(1.83)^2
3.61/(6.18)^2 = .09 nope
try -11.4
.016 = 1.9/11.4^2
3.61/(225) = .016

So x = -11.4