To find the net force on Q3, we need to calculate the force exerted on Q3 by Q1 and Q2 separately, and then add the two forces together using vector addition.
The magnitude of the force between two charges can be calculated using Coulomb's law: F = k * |Q1 * Q2| / r^2, where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q1 and Q2 are the magnitudes of the charges, and r is the distance between the charges.
1. Force between Q1 and Q3:
Distance between Q1 and Q3, r1 = sqrt((0.4m)^2 + (0.3m)^2) = 0.5m
Magnitude of force, F1 = (8.99 x 10^9) * |2 * 5| / (0.5)^2 = 179.8 N
2. Force between Q2 and Q3:
Distance between Q2 and Q3, r2 = sqrt((0.4m)^2 + (0.3m)^2) = 0.5m
Magnitude of force, F2 = (8.99 x 10^9) * |4 * 5| / (0.5)^2 = 359.6 N
Now, we need to find the directions of the forces. The force between like charges (positive and positive) is repulsive, so both F1 and F2 will act away from Q3.
To find the net force on Q3, we need to calculate the x and y components of each force:
F1x = F1 * cos(45°) = 179.8 * cos(45°) = 127 N
F1y = F1 * sin(45°) = 179.8 * sin(45°) = 127 N
F2x = F2 * cos(45°) = 359.6 * cos(45°) = 254 N
F2y = F2 * sin(45°) = 359.6 * sin(45°) = 254 N
Now, we can add the x and y components of the forces:
Net force in the x-direction = F1x + F2x = 127 + 254 = 381 N
Net force in the y-direction = F1y + F2y = 127 + 254 = 381 N
The magnitude of the net force on Q3 can be calculated using Pythagorean theorem:
Net force = sqrt((381)^2 + (381)^2) = sqrt(145,161) ≈ 381 N
The direction of the net force can be calculated using trigonometry:
tanθ = (F1y + F2y) / (F1x + F2x) = 381 / 381 = 1
θ = tan^(-1)(1) = 45°
Therefore, the magnitude of the net force on Q3 is approximately 381 N, and the direction of the net force is 45° with respect to the positive x-axis.
Three charges are placed on the xy _ plane Q1 = 2uc at (0,0.3m),Q2= 4uc at (0,_0.3m)and Q3= 5uc at (0.4m,0) what is the magnitude and direction the net forces on Q3
1 answer