Answer:
Step-by-step explanation:
We can use Coulomb's law to calculate the force between each pair of charges, and then use vector addition to find the net force on q2.
The distance between q2 and q1 is the same as the distance between q2 and q3, so we only need to calculate the force between q2 and one of the other charges and then multiply by 2.
The magnitude of the force between two charges q1 and q2 separated by a distance r is given by Coulomb's law:
F = k * |q1| * |q2| / r^2
where k is Coulomb's constant (9.0 x 10^9 N m^2 / C^2). Note that we take the absolute values of the charges since the forces are repulsive between like charges and attractive between opposite charges.
The direction of the force is along the line connecting the two charges, and is attractive if the charges are opposite and repulsive if the charges are the same.
Using this formula, we find:
The force on q2 due to q1 has magnitude |F1| = k * |q1| * |q2| / r^2 = (9.0 x 10^9 N m^2 / C^2) * (50 x 10^-9 C) * (50 x 10^-9 C) / (0.05 m)^2 = 9 N
The force on q2 due to q3 has magnitude |F3| = k * |q3| * |q2| / r^2 = (9.0 x 10^9 N m^2 / C^2) * (30 x 10^-9 C) * (50 x 10^-9 C) / (0.1 m)^2 = 1.35 N
To find the net force on q2, we need to add these two forces as vectors. Since the two forces are along the same line, we can simply add their magnitudes and determine the direction based on whether the charges are opposite or the same.
If the charges are opposite, the net force is attractive and points toward the other charge; if the charges are the same, the net force is repulsive and points away from the other charge.
In this case, q1 and q2 have opposite charges, so the force on q2 due to q1 points toward q1. Similarly, q2 and q3 have the same charge, so the force on q2 due to q3 points away from q3. Therefore, the net force on q2 is:
Fnet = F1 - F3 = 9 N - 1.35 N = 7.65 N
To find the direction of the net force, we can use the arctangent function:
θ = arctan(F3 / F1) = arctan(1.35 N / 9 N) = 8.4°
Since q1 and q2 are in the second quadrant and q3 is in the fourth quadrant, the net force on q2 points in the second quadrant.
So the net force on q2 has magnitude 7.65 N and direction 181.6° (counterclockwise from the negative x-axis).
To express the net force in component form, we can use trigonometry:
Fx = Fnet * cos(θ) = 7.65 N * cos(8.4°) = 7.63 N
Fy = Fnet * sin(θ) = 7.65 N * sin(8.4°) = 1.11 N
Therefore, the net force on q2 has x-component 7.63 N