Question

Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are Q1= 6.7 μC, Q2 = -9.4 μC, Q3= -4.9 μC .

1. Calculate the magnitude of the net force on particle 2 due to the other two.
2. Calculate the magnitude of the net force on particle 3 due to the other two.
3. Calculate the direction of the net force on particle 3 due to the other two.

Answer 1

1) Let charge $Q_1$ placed at the top vertice of the equilateral triangle and charges $Q_2$ and $Q_3$ placed in the corners of the base of the equilateral triangle. Let the side $Q_1Q_2$ be the $x$-axis. The net force on charge $Q_2$ is the vector sum of forces $F_{12}$ and $F_{32}$ due to charges $Q_1$ and $Q_3$.

Let’s find the magnitudes of these forces:

$$F_{12}=\dfrac{k|Q_1Q_2|}{r^2},$$ $$F_{12}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|6.7\cdot10^{-6}\ C\cdot(-9.4\cdot10^{-6}\ C)|}{(1.2\ m)^2}=0.394\ N,$$ $$F_{32}=\dfrac{k|Q_3Q_2|}{r^2},$$ $$F_{32}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|(-4.9\cdot10^{-6}\ C)\cdot(-9.4\cdot10^{-6}\ C)|}{(1.2\ m)^2}=0.288\ N.$$

Then, we can find the projections of forces $F_{12}$ and $F_{32}$ on axis $x$ and $y$:

$$F_x=F_{12x}+F_{32x}=-0.394\ N-0.288\ N\cdot cos60^{\circ}=-0.538 N,$$ $$F_y=F_{12y}+F_{32y}=0-0.288\ N\cdot sin60^{\circ}=-0.249 N.$$

Finally, the net electric force can be found from the Pythagorean theorem:

$$F_{net}=\sqrt{F_x^2+F_y^2},$$ $$F_{net}=\sqrt{(-0.538\ N)^2+(-0.249\ N)^2}=0.593\ N.$$

2) Let the side $Q_1Q_3$ be the $x$-axis. The net force on charge $Q_3$ is the vector sum of forces $F_{13}$ and $F_{23}$ due to charges $Q_1$ and $Q_2$.

Let’s find the magnitudes of these forces:

$$F_{13}=\dfrac{k|Q_1Q_3|}{r^2},$$ $$F_{13}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|6.7\cdot10^{-6}\ C\cdot(-4.9\cdot10^{-6}\ C)|}{(1.2\ m)^2}=0.205\ N,$$ $$F_{23}=\dfrac{k|Q_2Q_3|}{r^2},$$ $$F_{23}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot|(-9.4\cdot10^{-6}\ C)\cdot(-4.9\cdot10^{-6}\ C)|}{(1.2\ m)^2}=0.288\ N.$$

Then, we can find the projections of forces $F_{13}$ and $F_{23}$ on axis $x$ and $y$:

$$F_x=F_{13x}+F_{23x}=-0.205\ N+0.288\ N\cdot cos60^{\circ}=-0.061 N,$$ $$F_y=F_{13y}+F_{23y}=0-0.288\ N\cdot sin60^{\circ}=-0.249 N.$$

Finally, the net electric force can be found from the Pythagorean theorem:

$$F_{net}=\sqrt{F_x^2+F_y^2},$$ $$F_{net}=\sqrt{(-0.061\ N)^2+(-0.249\ N)^2}=0.256\ N.$$

3) We can find the direction of the net force on charge $Q_3$ as follows:

$$cos\theta=\dfrac{F_x}{F_{net}},$$ $$\theta=cos^{-1}(\dfrac{F_x}{F_{net}})=cos^{-1}(\dfrac{-0.061 N}{0.256\ N})=104^{\circ}.$$