Question

Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 2.20 µC, and L =0.810 m). Calculate the total electric force on the 7.00-µC charge.

magnitude	N
direction	° (counterclockwise from the +x axis)

Answer 1

The force on charge 3 due to charge 1 is given as,

$$ \vec{F}_{31}=\frac{k q_{1} q_{3}}{L^{2}}(\cos \theta \hat{i}+\sin \theta \hat{j}) $$

The force on charge 3 due to charge 2 is given as, \(\vec{F}_{32}=\frac{k q_{2} q_{3}}{L^{2}}(\cos \theta \hat{i}-\sin \theta \hat{j})\)

The net force on charge 3 is given as, \(\vec{F}=\vec{F}_{31}+\vec{F}_{32}\)

$$ \begin{array}{l} =\frac{k q_{1} q_{3}}{L^{2}}(\cos \theta \hat{i}+\sin \theta \hat{j})+\frac{k q_{2} q_{3}}{L^{2}}(\cos \theta \hat{i}-\sin \theta \hat{j}) \\ =\frac{k q_{3}}{L^{2}}\left(\left(q_{1}+q_{2}\right) \cos \theta \hat{i}+\left(q_{1}-q_{2}\right) \sin \theta \hat{j}\right) \\ =\frac{\left(8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(7.00 \times 10^{-6} \mathrm{C}\right)}{(0.810 \mathrm{~m})^{2}}((2.2+4.0) \cos 60 \hat{i}+(2.2-4.0) \sin 60 \hat{j})\left(10^{-6} \mathrm{C}\right) \\ =(0.2973 \mathrm{~N}) \hat{i}-(0.1495 \mathrm{~N}) \hat{j} \end{array} $$

The magnitude of the force is, \(F=\sqrt{F_{x}^{2}+F_{y}^{2}}\)

$$ \begin{array}{l} =\sqrt{(0.2973 \mathrm{~N})^{2}+(-0.1495 \mathrm{~N})^{2}} \\ =0.3327 \mathrm{~N} \\ \approx 0.333 \mathrm{~N} \end{array} $$

The direction of the force below the horizontal axis is, \(\phi=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)\)

$$ \begin{array}{l} =\tan ^{-1}\left(\frac{-0.1495}{0.2973}\right) \\ =-26.7^{\circ} \end{array} $$

The direction of the force counterclockwise from the positive \(x\) axis is given as, \(\phi=360^{\circ}-26.7^{\circ}\)

\(=333^{\circ}\)