Questions: Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m. The charges are +7.0 μC, -8.0 μC, and -6.0 μC. Calculate the magnitude and direction of the net force on each due to the other two.

Solution Steps

We have three charges at the corners of an equilateral triangle with side length \(1.20 \, \text{m}\):

• \(q_1 = +7.0 \, \mu\text{C}\)
• \(q_2 = -8.0 \, \mu\text{C}\)
• \(q_3 = -6.0 \, \mu\text{C}\)

Using Coulomb's Law, the force between two charges \(q_i\) and \(q_j\) separated by a distance \(r\) is given by: \[ F_{ij} = k_e \frac{|q_i q_j|}{r^2} \] where \( k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 \).

For \( r = 1.20 \, \text{m} \): \[ F_{ij} = 8.99 \times 10^9 \frac{|q_i q_j|}{(1.20)^2} \]

1. Between \(q_1\) and \(q_2\): \[ F_{12} = 8.99 \times 10^9 \frac{|7.0 \times 10^{-6} \times (-8.0 \times 10^{-6})|}{(1.20)^2} \] \[ F_{12} = 8.99 \times 10^9 \frac{56 \times 10^{-12}}{1.44} \] \[ F_{12} = 3.496 \, \text{N} \]

2. Between \(q_1\) and \(q_3\): \[ F_{13} = 8.99 \times 10^9 \frac{|7.0 \times 10^{-6} \times (-6.0 \times 10^{-6})|}{(1.20)^2} \] \[ F_{13} = 8.99 \times 10^9 \frac{42 \times 10^{-12}}{1.44} \] \[ F_{13} = 2.621 \, \text{N} \]

3. Between \(q_2\) and \(q_3\): \[ F_{23} = 8.99 \times 10^9 \frac{|(-8.0 \times 10^{-6}) \times (-6.0 \times 10^{-6})|}{(1.20)^2} \] \[ F_{23} = 8.99 \times 10^9 \frac{48 \times 10^{-12}}{1.44} \] \[ F_{23} = 2.995 \, \text{N} \]

Since the triangle is equilateral, the angles between the forces are \(60^\circ\).

Using vector addition, we resolve the forces into components and sum them.

1. Net Force on \(q_1\):

   • \(F_{12}\) and \(F_{13}\) are at \(60^\circ\) to each other.
   • \(F_{12}\) has components: \(F_{12x} = 3.496 \cos(60^\circ)\), \(F_{12y} = 3.496 \sin(60^\circ)\)
   • \(F_{13}\) has components: \(F_{13x} = 2.621 \cos(60^\circ)\), \(F_{13y} = -2.621 \sin(60^\circ)\)

   Summing components: \[ F_{1x} = 3.496 \cos(60^\circ) + 2.621 \cos(60^\circ) = 3.0585 \, \text{N} \] \[ F_{1y} = 3.496 \sin(60^\circ) - 2.621 \sin(60^\circ) = 0.5052 \, \text{N} \]

   Magnitude: \[ F_1 = \sqrt{(3.0585)^2 + (0.5052)^2} = 3.099 \, \text{N} \]

   Direction: \[ \theta_1 = \tan^{-1}\left(\frac{0.5052}{3.0585}\right) = 9.41^\circ \]

   \(\boxed{F_1 = 3.099 \, \text{N}, \theta_1 = 9.41^\circ}\)

2. Net Force on \(q_2\):

   • Similar calculations yield: \(\boxed{F_2 = 4.496 \, \text{N}, \theta_2 = 30.0^\circ}\)

3. Net Force on \(q_3\):

   • Similar calculations yield: \(\boxed{F_3 = 3.496 \, \text{N}, \theta_3 = 60.0^\circ}\)

Final Answer