**
in cm
**
should a third charged be placed such that the net force on the third charge is zero?

The answer is -1.

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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.