Coulomb's law:

$\overline{){\mathbf{F}}{\mathbf{=}}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$, where k is the coulomb's constant and r is the separation between the two charges.

The magnitude of the net force:

$\overline{){{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}}{\mathbf{=}}\sqrt{{{\mathbf{F}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{F}}_{\mathbf{y}}}^{\mathbf{2}}}}$

k = 9 × 10^{9} N·m^{2}/C^{2}

The force due to charge Q_{1} is:

$\begin{array}{rcl}{\mathbf{F}}_{\mathbf{1}\mathbf{A}}& \mathbf{=}& \frac{\mathbf{k}{\mathbf{Q}}_{\mathbf{1}}{\mathbf{Q}}_{\mathbf{A}}}{{{\mathbf{r}}_{\mathbf{1}\mathbf{A}}}^{\mathbf{2}}}\\ & \mathbf{=}& \frac{\mathbf{(}\mathbf{9}\mathbf{\times}{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\mathbf{(}\mathbf{21}\mathbf{.}\mathbf{4}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}\mathbf{(}\mathbf{13}\mathbf{.}\mathbf{5}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}}{{\mathbf{(}\mathbf{80}\mathbf{.}\mathbf{1}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}\end{array}$

Pictured on the right are three-point charges Q_{1} = 21.4 μC, Q_{2} = -30.6 μC, and Q_{3 }= 57.3 μC arranged according to the figure on the right. A fourth point charge is located at point A with a charge of Q_{A} = 13.5 μC. Calculate the magnitude of the net force on the charge at point A. The answer should be in N.

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