Two charges of 2.0 × 10^ -6 C and -1.0 × 10^ -6 C are placed at a...

Let $q_1 = 2.0 \times 10^{-6} \text{ C}$ and $q_2 = -1.0 \times 10^{-6} \text{ C}$.

The distance between the charges is $d = 10 \text{ cm}$.

For the net force on a third charge to be zero, the net electric field at that point must be zero. Since the charges have opposite signs, the neutral point lies on the line joining the charges, outside the region between them, and closer to the charge with the smaller magnitude ($q_2$).

Let the neutral point be at a distance $x$ from $q_1$ on the side of $q_2$. The distance of this point from $q_2$ is $x - 10$.

Equating the magnitudes of the electric fields produced by $q_1$ and $q_2$ at this point:

$\dfrac{k |q_1|}{x^2} = \dfrac{k |q_2|}{(x - 10)^2}$

$\dfrac{2.0 \times 10^{-6}}{x^2} = \dfrac{1.0 \times 10^{-6}}{(x - 10)^2}$

$\dfrac{2}{x^2} = \dfrac{1}{(x - 10)^2}$

Taking the square root of both sides:

$\dfrac{\sqrt{2}}{x} = \dfrac{1}{x - 10}$

$\sqrt{2} (x - 10) = x$

$x (\sqrt{2} - 1) = 10 \sqrt{2}$

$x = \dfrac{10 \sqrt{2}}{\sqrt{2} - 1}$

Rationalizing the denominator:

$x = 10 \sqrt{2} (\sqrt{2} + 1) = 10 (2 + \sqrt{2})$

Using $\sqrt{2} \approx 1.414$:

$x = 10 (2 + 1.414) = 34.14 \text{ cm}$

The position is $34.1 \text{ cm}$ from the larger magnitude charge on the side of the smaller magnitude charge.