# Question #a20f5

Jul 29, 2016

$\frac{1}{3}$ meters to the right of $q 1$ and $3$ meters to the left of ${q}_{1}$ (assuming ${q}_{2}$ is to the right of ${q}_{1}$)

#### Explanation:

Electric field from ${q}_{1}$ to cancel ${q}_{2}$ would mean:
${E}_{1} = {E}_{2}$

$k {q}_{1} / {r}^{2} = k {q}_{2} / {\left(1 - r\right)}^{2}$
$r$ is the distance from ${q}_{1}$.

Rearrange the equation and you have a quadratic equation:
$3 {r}^{2} + 2 r - 1 = 0$

Find the roots and you have $\frac{1}{3} m$ (to the right) and $- 3 m$ (to the left).