Question

Three charged particles are located at the corners of an equilateral triangle as shown in the figure below (let q = 3.20 µC, and L = 0.650 m). Calculate the total electric force on the 7.00-µC charge.

Answer 1

The electric force on charge \(7.00 \mu \mathrm{C}\) due to charge \(3.20 \mu \mathrm{C}\) is

$$ \begin{aligned} F_{1} &=\frac{k c_{1} q_{2}}{L^{2}} \\ &=\frac{\left(9 \times 10^{9}\right)\left(7.00 \times 10^{-6}\right)\left(3.20 \times 10^{-6}\right)}{(0.650)^{2}}=0.477 \mathrm{~N} \end{aligned} $$

The electric force on charge \(7.00 \mu \mathrm{C}\) due to charge \(3.20 \mu \mathrm{C}\) is

$$ \begin{aligned} F_{2} &=\frac{k q_{1} q_{3}}{L^{2}} \\ &=\frac{\left(9 \times 10^{9}\right)\left(7.00 \times 10^{-6}\right)\left(4.00 \times 10^{-6}\right)}{(0.650)^{2}}=0.596 \mathrm{~N} \end{aligned} $$

The net electric force on charge \(7.00 \mu \mathrm{C}\) along \(\mathrm{x}\) -direction is

$$ \begin{aligned} F_{x} &=\left(F_{1}+F_{2}\right) \cos 60^{\circ} \\ &=(0.477 \mathrm{~N}+0.596 \mathrm{~N}) \cos 60^{\circ}=0.5365 \mathrm{~N} \end{aligned} $$

The net electric force on charge \(7.00 \mu \mathrm{C}\) along y-direction is

$$ \begin{aligned} F_{y} &=\left(F_{1}-F_{2}\right) \sin 60^{\circ} \\ &=(0.477 \mathrm{~N}-0.596 \mathrm{~N}) \sin 60^{\circ}=-0.103057 \mathrm{~N} \end{aligned} $$

The magnitude of net electric force is

$$ \begin{aligned} F &=\sqrt{F_{x}^{2}+F_{y}^{2}} \\ &=\sqrt{(0.5365 \mathrm{~N})^{2}+(-0.103057 \mathrm{~N})^{2}}=0.546 \mathrm{~N} \end{aligned} $$

Direction is

$$ \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=\tan ^{-1}\left(\frac{-0.103057 \mathrm{~N}}{0.5365 \mathrm{~N}}\right)=-10.87^{\circ}=349^{\circ} $$

from the positive \(x\) -axis.