Questions: As shown in the figure, three charges are at the vertices of an equilateral triangle of side 5 cm. The charge q₁ is -6.5 nC, charge q₂ is -2.5 nC, and charge q₃ is -2.5 nC. What is the magnitude and direction (angle made with positive x-direction) of the net electric force on the charge q₁ due to the other two charges? ( k = 1 / 4 πε₀ = 9.0 × 10⁹ N·m² / C². Magnitude = .00000858 N Direction = 30 degree

Solution Steps

The force F₂₁ due to q2 on q1 is given by Coulomb's law: F₂₁ = k * |q1 * q2| / r² F₂₁ = (9.0 x 10⁹ Nm²/C²) * |-6.5 x 10^-9 C * -2.5 x 10^-9 C| / (0.05 m)² F₂₁ = 5.85 x 10^-5 N, directed towards q2 (along the line joining q1 and q2).

Similarly, the force F₃₁ due to q3 on q1 is: F₃₁ = k * |q1 * q3| / r² F₃₁ = (9.0 x 10⁹ Nm²/C²) * |-6.5 x 10^-9 C * -2.5 x 10^-9 C| / (0.05 m)² F₃₁ = 5.85 x 10^-5 N, directed towards q3 (along the line joining q1 and q3).

Since the triangle is equilateral, the angle between F₂₁ and F₃₁ is 60°. The x-components of the two forces add up, while the y-components cancel out due to symmetry. The magnitude of the net force F_net is: F_net = 2 * F₂₁ * cos(30°) = 2 * 5.85 x 10^-5 N * cos(30°) = 1.013 x 10^-4 N. The net force is directed along the positive x-axis (bisecting the angle between F₂₁ and F₃₁). Therefore, the angle it makes with the positive x-direction is 0°.

Final Answer