Three identical charges, each of 2.0 × 10^ -6 C , are held fixed at...

Let the charges be placed at the vertices of the equilateral triangle. The force exerted by one charge on another at a distance $a$ is given by Coulomb's law:

$F = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{a^2}$

Substituting the given values $q = 2.0 \times 10^{-6} \text{ C}$ and $a = 5 \times 10^{-2} \text{ m}$:

$F = \dfrac{9 \times 10^9 \times (2.0 \times 10^{-6})^2}{(5 \times 10^{-2})^2}$

$F = \dfrac{9 \times 10^9 \times 4.0 \times 10^{-12}}{25 \times 10^{-4}} = \dfrac{36 \times 10^{-3}}{25 \times 10^{-4}} = 14.4 \text{ N}$

The charge at any vertex experiences two such forces from the other two charges. Since the triangle is equilateral, the angle between these two forces is $60^{{}^{{}^{\circ}}}$.

The magnitude of the resultant force is:

$F_{net} = \sqrt{F^2 + F^2 + 2F^2 \cos 60^{{}^{{}^{\circ}}}}$

$F_{net} = \sqrt{2F^2 + 2F^2 \left(\dfrac{1}{2}\right)} = \sqrt{3F^2} = F\sqrt{3}$

$F_{net} = 14.4 \times \sqrt{3} \approx 14.4 \times 1.732 \approx 24.9 \text{ N}$