Answer
$q = -1.5~nC$
Work Step by Step
Let $q_1$ be the charge at the origin. Then $r_1 = 1.0~m$ and $r_2 = 0.50~m$. Note that $r_1 = 2~r_2$
At the point $x = 1.0~m$, the sum of the electric fields due to the two point charges is zero. We can find $q$:
$E = \frac{k~q_1}{r_1^2} + \frac{k~q}{r_2^2} = 0$
$\frac{k~q}{r_2^2} = -\frac{k~q_1}{r_1^2}$
$\frac{q}{r_2^2} = -\frac{q_1}{r_1^2}$
$q = -\frac{q_1~r_2^2}{r_1^2}$
$q = -\frac{q_1~r_2^2}{(2~r_2)^2}$
$q = -\frac{q_1}{4}$
$q = -\frac{+6.0~nC}{4}$
$q = -1.5~nC$
