Three charged particles are located at the vertices of a right angled triangle. Charge A(5.0*10^-6) has Cartesian coordinates of (0,4); Charge B(-5.0*10^-6) is at the origin; Charge C(4.0*10^-6) has coordinates (5,0),where the coordinates are in meters. What is the net force on each charge?

The net force on Charge A is (0, 0).

The net force on Charge B is (0, -9.6 x 10^-6 N).

The net force on Charge C is (9.6 x 10^-6 N, 0).

To find the net force on each charge, we need to calculate the force between each pair of charges using Coulomb's Law and then add up the forces.

Let's start with Charge A:

1. Calculate the force between Charge A and Charge B:
- Distance between A and B: d_AB = √((x_B - x_A)^2 + (y_B - y_A)^2)
= √((0 - 0)^2 + (0 - 4)^2)
= √(0 + 16)
= 4.0 meters

- Force between A and B: F_AB = k * (q_A * q_B) / d_AB^2
- Where k is the electrostatic constant (k = 9.0 * 10^9 Nm^2/C^2)
- q_A and q_B are the charges of A and B respectively
- Plugging in the values: F_AB = (9.0 * 10^9) * (-5.0 * 10^-6 * 5.0 * 10^-6) / (4.0)^2
= -56.25 N (Note the negative sign indicates an attractive force)

2. Calculate the force between Charge A and Charge C:
- Distance between A and C: d_AC = √((x_C - x_A)^2 + (y_C - y_A)^2)
= √((5 - 0)^2 + (0 - 4)^2)
= √(25 + 16)
= √41.0 meters

- Force between A and C: F_AC = k * (q_A * q_C) / d_AC^2
- Plugging in the values: F_AC = (9.0 * 10^9) * (5.0 * 10^-6 * 4.0 * 10^-6) / (41.0)^2
= 1.50 N

Now, let's move on to Charge B:

3. Calculate the force between Charge B and Charge C:
- Distance between B and C: d_BC = √((x_C - x_B)^2 + (y_C - y_B)^2)
= √((5 - 0)^2 + (0 - 0)^2)
= √(25 + 0)
= 5.0 meters

- Force between B and C: F_BC = k * (q_B * q_C) / d_BC^2
- Plugging in the values: F_BC = (9.0 * 10^9) * (-5.0 * 10^-6 * 4.0 * 10^-6) / (5.0)^2
= -14.4 N (Note the negative sign indicates an attractive force)

Finally, let's calculate the net force on each charge by adding up the forces:

Net force on Charge A = F_AB + F_AC
= -56.25 N + 1.50 N
= -54.75 N

Net force on Charge B = F_AB + F_BC
= -56.25 N - 14.4 N
= -70.65 N

Net force on Charge C = F_AC + F_BC
= 1.50 N - 14.4 N
= -12.9 N

Therefore, the net force on Charge A is -54.75 N, on Charge B is -70.65 N, and on Charge C is -12.9 N.

To find the net force on each charged particle, we need to calculate the force between each pair of charged particles using Coulomb's Law, and then add up the forces vectorially.

Coulomb's Law states that the force between two charged particles is given by the equation:

F = (k * |q1 * q2|) / r^2

Where:
- F is the force between the particles
- k is the electrostatic constant, approximately equal to 9.0 * 10^9 Nm^2/C^2
- q1 and q2 are the magnitudes of the charges on the two particles
- r is the distance between the particles

Let's calculate the net force on each charge:

1. Force on Charge A:
- Distance between A and B: AB = √((x2 - x1)^2 + (y2 - y1)^2)
AB = √((0 - 0)^2 + (0 - 4)^2) = √16 = 4 m
- Force between A and B: FAB = (k * |qA * qB|) / AB^2
FAB = (9.0 * 10^9 Nm^2/C^2 * |5.0 * 10^-6 C * -5.0 * 10^-6 C|) / (4 m)^2
FAB = (9.0 * 10^9 * 25.0 * 10^-12) / 16
FAB ≈ 14.1 * 10^-3 N (approximating to three significant figures)

2. Force on Charge B:
- Force between B and A is equal in magnitude, but opposite in direction to FAB.
FBA = -14.1 * 10^-3 N

3. Force on Charge C:
- Distance between C and B: CB = √((x2 - x1)^2 + (y2 - y1)^2)
CB = √((5 - 0)^2 + (0 - 0)^2) = √25 = 5 m
- Force between C and B: FCB = (k * |qC * qB|) / CB^2
FCB = (9.0 * 10^9 Nm^2/C^2 * |4.0 * 10^-6 C * -5.0 * 10^-6 C|) / (5 m)^2
FCB = (9.0 * 10^9 * 20.0 * 10^-12) / 25
FCB ≈ 7.2 * 10^-3 N (approximating to three significant figures)

4. Force on Charge B (from C) is equal in magnitude, but opposite in direction to FCB.
FBC = -7.2 * 10^-3 N

5. Net force on Charge A:
- Since the force between Charge A and Charge B are in opposite directions, the net force is the vector sum of the forces:
FA_net = √(FAB^2 + FBC^2)
FA_net = √((14.1 * 10^-3)^2 + 0) (since FBC and FAB are perpendicular forces)
FA_net ≈ 14.1 * 10^-3 N (approximating to three significant figures)

6. Net force on Charge B:
- Since the force between Charge A and Charge B are equal in magnitude but opposite in direction, and the force between Charge B and Charge C are equal in magnitude but opposite in direction, the net force is zero on Charge B:
FB_net = 0 N

7. Net force on Charge C:
- Since the force between Charge B and Charge C are in opposite directions, the net force is the vector sum of the forces:
FC_net = √(FBC^2 + FCB^2)
FC_net = √((7.2 * 10^-3)^2 + 0) (since FBC and FCB are perpendicular forces)
FC_net ≈ 7.2 * 10^-3 N (approximating to three significant figures)

In summary, the net forces on each charge are approximately:
- Net force on Charge A: 14.1 * 10^-3 N
- Net force on Charge B: 0 N
- Net force on Charge C: 7.2 * 10^-3 N