Vanishing Electric Field

$\vec{E_{net}}=\vec{E_{9 C}}+\vec{E_{1 C}}=0$

$\vec{E_{net}}=\dfrac{kQ_{9 C}}{r^2}\hat{r}_{9 C} + \dfrac{kQ_{1 C}}{r^2}\hat{r}_{1 C}=0$

Since the electric field points away from positive charges, if we choose an arbitrary point x in between the two charges, the electric field from the 9 C charge points in the positive x direction, while the electric field due to the 1 C charge points in the negative x direction at that point.

$\dfrac{kQ_{9 C}}{r^2}\hat{x} + \dfrac{kQ_{1 C}}{r^2}(-\hat{x})=0$

$\dfrac{kQ_{9 C}}{r^2}\hat{x} = \dfrac{kQ_{1 C}}{r^2}(\hat{x})$

$\dfrac{Q_{9 C}}{x^2} = \dfrac{Q_{1 C}}{(1-x)^2}$

$\dfrac{9}{x^2} = \dfrac{1}{(1-x)^2}$

x = 0.75 m or 1.5 m

We must be in the region where the electric fields from the two charges point in opposite directions so that they may cancel out. Since the charges have the same sign, we must be between the two charges , so the correct answer is x = 0.75 m

The electric field is proportional to charge and inverse squared distance: $E \propto q/r^2$ .

Thus the total electric field is zero at a points where $\frac{q_1}{r_1^2} = \frac{q_2}{r_2^2}\ \ \therefore\ \ \left(\frac{r_2}{r_1}\right)^2 = \frac{q_1}{q_2}.$ Plugging in what we know, we find that the point in question is $1/\sqrt{9} = 1/3$ closer to the second charge than to the first, giving the result $\boxed{0.75}$ m.