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

Let the third charge $q$ be placed at a distance $x$ from the larger charge $q_1 = 2.0 \times 10^{-6} \text{ C}$.

The distance from the smaller charge $q_2 = 1.0 \times 10^{-6} \text{ C}$ will be $(10 - x) \text{ cm}$.

For the net force on $q$ to be zero, the electrostatic forces due to $q_1$ and $q_2$ must be equal in magnitude and opposite in direction:

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

Substituting the given values:

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

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

Taking the square root of both sides:

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

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

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

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

Rationalizing the denominator:

$x = 10\sqrt{2}(\sqrt{2} - 1) = 20 - 10\sqrt{2}$

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

$x = 20 - 14.14 = 5.86 \text{ cm}$

Rounding to one decimal place, we get $x \approx 5.9 \text{ cm}$ from the larger charge.