A +5.8 µC charge is placed at the origin and a -3.9 µC charge is placed at x = 25 cm?

Question

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

Answer 1

let q1 = 5.8 µC, q2 = 3.9 µC, d = 25 cm
since q1 is +, q2 is -, a third charge q will experience an attractive force and a repulsive force, if these 2 forces have same magnitude and opposite directions, the net force will be zero.
On X-axis, there are 3 regions: left of q1, between q1 and q2, right of q2.
On the left of q1: the above 2 forces are opposite, but note that q1 > q2, the distance between q1 and q is shorter then the distance between q2 and q, so the force due to q1 is much larger than the force due to q2, these 2 forces can't be same magnitude. It means the third charge q can't be put at left of q1 to get zero net force.
If q is put between q1 and q2, the 2 forces have same direction, they can't cancel each other, the net force can't be zero.
If q is put right of q2, the net force can be zero. Let q is at A, the distance between q and q2 is a. Then the magnitude of a force is kq*q2/a^2, the other is kq*q1/(d+a)^2, we have
kq*q2/a^2 = kq*q1/(d+a)^2
q2/a^2 = q1/(d+a)^2
(d+a)^2/a^2 = q1/q2
(d+a)/a = sqrt(q1/q2)
d/a = sqrt(q1/q2) - 1
a = d/[sqrt(q1/q2) - 1] = 25/[sqrt(5.8/3.9)-1] = 114 cm
so the coordinate where q is placed is d+a = 139 cm

Answer 2

In theory, this is easy.

Whatever the third charge, it will feel an attractive force from one and a repulsive force from the other. So for it to feel no net force, the two forces have to cancel.

That means they must be along the same line, which means the third charge has to be on the X axis and we can use scalars instead of vectors.

Beyond that, if the third charge is q at location x, then
F1 = (q q1)/(x1 - x)^2
F2 = (q q2)/(x2 - x)^2) and
F1 = -F2

You know q1, x1, q2, and x2; the exact value of q should not matter (as long as it isn't 0) because both forces are directly proportional to it. So you have everything you need to solve for x.

Of course, the real answer is nowhere - any solution you get is in unstable equilibrium - but that isn't the desired answer.