Force,

$\overline{)\begin{array}{rcl}{\mathbf{F}}& {\mathbf{=}}& \frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\end{array}}$

q_{1} =3 mC = 3 × 10^{-6} C

q_{2} = -5 mC = -5 × 10^{-6} C

q_{3} = -8.00 mC = -8 × 10^{-6} C

x_{12} = 0.2m

q_{1} is attracted by both q_{2} and q_{3}. F_{net} is directed to the left (negative) while F_{12} is directed to the right (positive).

$\begin{array}{rcl}{\mathbf{F}}_{\mathbf{1}\mathbf{,}\mathbf{}\mathbf{net}}& \mathbf{=}& {\mathbf{F}}_{\mathbf{12}}\mathbf{+}{\mathbf{F}}_{\mathbf{13}}\\ {\mathbf{F}}_{\mathbf{13}}& \mathbf{=}& {\mathbf{F}}_{\mathbf{1}\mathbf{,}\mathbf{}\mathbf{net}}\mathbf{-}{\mathbf{F}}_{\mathbf{12}}\\ & \mathbf{=}& \mathbf{-}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{-}\frac{{\mathbf{kq}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{{\mathbf{r}}_{\mathbf{12}}}^{\mathbf{2}}}\\ & \mathbf{=}& \mathbf{-}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{-}\frac{\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{9}}\mathbf{)}\mathbf{(}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{)}}{\mathbf{0}\mathbf{.}{\mathbf{2}}^{\mathbf{2}}}\\ & \mathbf{=}& \mathbf{-}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{375}\\ & \mathbf{=}& \mathbf{-}\mathbf{10}\mathbf{.}\mathbf{37}\end{array}$

Three point charges are arranged along the x-axis. Charge q_{1} = 3 mc is at the origin and charge q_{2} = -5 mc is at x = 0.2 m. Charge q_{3} = -8.00 mc.

Where is q_{3} located if the net force on q_{1} is 7.00 N in the -x-direction?

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