Balancing The Forces

We solve this using Coulomb's law: $F=k\frac{q_1q_2}{r^2}$ . Plugging in, we get that: $k\frac{1*q}{(x+3)^2} = k\frac{4*q}{(x-3)^2}$ , where q is the charge on the new object. We cancel the q's and the k's and multiply through by the denominators to obtain: $4x^2 + 24x+3.6 = x^2-6x+9$ . Thus $3x^2 +30x +27 = 0$ , $(3x+3)(x+9) = 0$ and we find that the places where the forces from the two charges are equal magnitude are -1 and -9. Clearly -9 is extraneous, so x = -1.

Let the charge of the third object be Q

Let the distance between -3cm and the position of Q be r

Total distance between the 3cm & -3cm = 3 - (-3) = 6cm.

Electrostatic Force of Attraction/Repulsion from -3cm = (k x 1 x Q) / (r^2);

Electrostatic Force of Attraction/Repulsion from 3cm = (k x 4 x Q) / ((6 - r)^2);

Since the net force must be zero, it follows that these to forces must be equated.

Thus, by equating, we cancel out k constant and Q. We are left with the quadratic:

-3(r^2) -12(r) + 36 = 0

Solving this equation, we achieve r = +2 or -6

Since displacement cannot be negative, r = +2

Thus, -3cm + 2cm = -1cm