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Two point charges are located on the \(x\) -axis of a coordinate system:
\(q_{1}=1.0 \mathrm{nC}\) is at \(x=+2.0 \mathrm{cm},\) and \(q_{2}=-3.0 \mathrm{nC}\) is at \(x=\)\(+4.0 \mathrm{cm} .\) What is the total electric force exerted by \(q_{1}\) and \(q_{2}\) on a
charge \(q_{3}=5.0 \mathrm{nC}\) at \(x=0 ?\)

\( \overrightarrow{F_{3}}?? \)

\({{\overrightarrow{F}}_{1 on 3}}??\)

\({{\overrightarrow{F}}_{2 on 3}}??\)

\( F_{1 on 3}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left|q, q_{3}\right|}{r_{13}^{2}} \)

\( F_{2 on}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\left|q_{2} q_{3}\right|}{r_{23}^{2}} \)

\( F_{1  on 3} =9 * 10^{9} * \frac{|\left(1*10^{-9}\right)\left(5*10^{-9}\right)|}{(0.02)^{2}} \)

\( =1.12 * 10^{-4} N \)

\( =112 M_{N} =\vec{F}_{\text { i on } 3} =-(112 \mathrm{M_N}) \hat{\mathfrak{i}} \)

\( F_{2 on 3}=9 * 10^{9} * \frac{\left|\left(-3 * 10^{-9}\right)\left(5 * 10^{-9}\right)\right|}{(0.04)^{2}} \)

\( =84 M_ N=(84 M_ N \hat{i}) \)

\( \vec{F}_{3}=\vec{F}_{1 on 3}+\vec{F}_{2 on 3} \)

\( =-112 \hat{\imath}+84 \hat{\imath}=-28 \hat{\imath}(M_N) \)

Three point charges are arranged on a line. Charge
\(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{cm} .\) Charge \(q_{1}\) is at \(x=+2.00 \mathrm{cm} .\) What is \(q_{1}\)
(magnitude and sign) if the net force on \(q_{3}\) is zero?

\( F_{2}={k} \frac{\left|q_{2} q_{3}\right|}{r_{2}^{2}} \)

\( =9 * 10^{9} * \frac{\left|-3 * 10^{-9} * 5 * 10^{-9}\right|}{\left(4 * 10^{-2}\right)^{2}}= \)

\( F_{2}=F_{1}=k \frac{|q_1 q_{3} |}{r_{1}^{2}} \)

\( F_{1}=F_{2} \longrightarrow k \frac{\left|q_{1}\right||q _ 3|}{r_{1}^{2}} =k \frac{|q_{2}| | q_{3} |}{r_{2}^{2}} \)

\( \longrightarrow \frac{|q_1 |}{r_{1}^{2}} \frac{\left|q_{2}\right|}{r_{2}^{2}} \rightarrow|q_1|= \left|q_{2}\right|\left(\frac{r_{1}}{r_{2}}\right)^{2} \)

\(= |-3 * 10^{-9}| (\frac{0.02}{0.04} )^{2} \)

\(∴ q_{1}=0.75 n C = 0.75 * 10^{-9} \mathrm{C} \)

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