Two point charges are located on the y axis as follows: charge q1 = -2.50nC at y1 = -0.700m , and charge q2 = 3.30nC at the origin (y = 0). What is the magnitude of the net force exerted by these two charges on a third charge q3 = 9.00nC located at y3 = 0.250m ?

What is the direction of the net force?

add them.

forcenet=F1 + F2
= -2.5*9/(.250+.7)^2 + 3.3*9/(.250+0)^2

put units on charges, I ignored that. Direction. think quickly, which force on q3 is greatest, the repulsive, or the attractive.

To find the magnitude of the net force exerted on the third charge, we can use the principle of superposition. The net force is the vector sum of the individual forces exerted by each of the charges.

1. Calculate the force exerted by q1 on q3 using Coulomb's Law:

F1 = k * |q1| * |q3| / r1^2

Where k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), |q1| and |q3| are the magnitudes of the charges, and r1 is the distance between them.

In this case, |q1| = 2.50 x 10^-9 C, |q3| = 9.00 x 10^-9 C, and r1 = y3 - y1 = 0.250m - (-0.700m) = 0.950m.

Plugging in the values:

F1 = (8.99 x 10^9 Nm^2/C^2) * (2.50 x 10^-9 C) * (9.00 x 10^-9 C) / (0.950m)^2

2. Calculate the force exerted by q2 on q3:

F2 = k * |q2| * |q3| / r2^2

Since q2 is located at the origin, the distance between q2 and q3 is simply y3 = 0.250m.

Plugging in the values:

F2 = (8.99 x 10^9 Nm^2/C^2) * (3.30 x 10^-9 C) * (9.00 x 10^-9 C) / (0.250m)^2

3. Calculate the net force by taking the vector sum of F1 and F2:

F_net = F1 + F2

Make sure to consider the directions and signs of the forces during the vector addition.

To determine the direction of the net force, we need to examine the signs and directions of each individual force. In this case:

- q1 is negative, so the force exerted by q1 on q3 is attractive.
- q2 is positive, so the force exerted by q2 on q3 is repulsive.

Since q2 is at the origin and q1 is above q3, both forces will act along the positive y-axis. The net force will be the vector sum of the magnitudes of the forces, pointing upwards along the positive y-axis.

To find the magnitude of the net force exerted by q1 and q2 on q3, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

1. Calculate the force exerted by q1 on q3:
- The distance between q1 and q3 is given by the y-coordinate difference:
r = |y1 - y3| = |-0.700m - 0.250m| = 0.950m
- Apply Coulomb's Law:
F1 = k * |q1| * |q3| / r^2, where k is the electrostatic constant (k = 9.0 x 10^9 N·m^2/C^2)
F1 = (9.0 x 10^9 N·m^2/C^2) * |(-2.50nC)| * |9.00nC| / (0.950m)^2

2. Calculate the force exerted by q2 on q3:
- The distance between q2 and q3 is given by the y-coordinate difference:
r = |y2 - y3| = |0 - 0.250m| = 0.250m
- Apply Coulomb's Law:
F2 = k * |q2| * |q3| / r^2, where k is the electrostatic constant (k = 9.0 x 10^9 N·m^2/C^2)
F2 = (9.0 x 10^9 N·m^2/C^2) * |(3.30nC)| * |9.00nC| / (0.250m)^2

3. Find the net force by summing the forces vectorially:
- The forces F1 and F2 are along the y-axis, and they have the same direction since both charges are positive.
- F_net = F1 + F2

4. Calculate the magnitude of the net force:
- F_net = √(F1^2 + F2^2)

5. Determine the direction of the net force:
- Since both charges q1 and q2 are positive and exert forces in the same direction on q3, the net force will also be in the positive y-direction.

Plug in the values and calculate F_net using the given charge values and distances.