Charge Q1=6.0 nC is at (0.30 m, 0), charge Q2=−1.0 nC is at (0, 0.10 m), and charge Q3=5.0 nC is at (0, 0). What is the magnitude of the net electrostatic force on the 5.0-nC charge due to the other charges?

What is the direction of the net electrostatic force on the 5.0-nC charge due to the other charges?

5.4*10^-6

Well, since we're dealing with charges, we can assume that they decided to have a charge party. Q1 is over there, dancing at (0.30 m, 0), Q2 is doing the worm at (0, 0.10 m), and Q3 is just chilling at (0, 0).

Now, let's find out how much of a force is being exerted on the 5.0-nC charge by these party animals.

First, we'll calculate the electrostatic force between Q1 and the 5.0-nC charge. We'll use Coulomb's Law, which states that the force between two charges is given by the equation F = k * (Q1 * Q2) / r^2, where k is the electrostatic constant, Q1 and Q2 are the charges, and r is the distance between them.

The force between Q1 and the 5.0-nC charge is given by F1 = (9 * 10^9 N m^2/C^2) * ((6.0 * 10^-9 C) * (5.0 * 10^-9 C)) / (0.30 m)^2.

Now, let's calculate the electrostatic force between Q2 and the 5.0-nC charge. This time, since Q2 is negative, it's like they're battling it out on the dance floor.

The force between Q2 and the 5.0-nC charge is given by F2 = (9 * 10^9 N m^2/C^2) * ((-1.0 * 10^-9 C) * (5.0 * 10^-9 C)) / (0.10 m)^2.

Finally, let's calculate the electrostatic force between Q3 and the 5.0-nC charge. Q3 might be chill, but they can still exert some force.

The force between Q3 and the 5.0-nC charge is given by F3 = (9 * 10^9 N m^2/C^2) * ((5.0 * 10^-9 C) * (5.0 * 10^-9 C)) / (0.10 m)^2.

To find the magnitude of the net electrostatic force on the 5.0-nC charge, we need to add up these three forces: net force = sqrt(F1^2 + F2^2 + F3^2).

As for the direction of the net electrostatic force, I would say it's probably telling the 5.0-nC charge to join the party and join the dance-off on the dance floor. It's all about having a good time, right? So, dance your way to the left or right, my friend!

Disclaimer: This response may contain some exaggerated humor for entertainment purposes. Please consult a physics textbook or a sane human for accurate calculations.

To find the magnitude of the net electrostatic force on the 5.0-nC charge due to the other charges, we can use Coulomb's law.

Coulomb's law states that the magnitude of the electrostatic force between two charges is given by the equation:

F = k * |Q1 * Q2| / r^2

where F is the electrostatic force, k is the electrostatic constant (approximately 8.99 x 10^9 N*m^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

Let's calculate the magnitude of the electrostatic force between the 5.0-nC charge and the other two charges separately.

For Q1:

Distance (r1) between Q1 and the 5.0-nC charge = 0.3 m
Charge of Q1 (Q1) = 6.0 nC = 6.0 x 10^-9 C

For Q2:

Distance (r2) between Q2 and the 5.0-nC charge = 0.1 m
Charge of Q2 (Q2) = -1.0 nC = -1.0 x 10^-9 C

Using Coulomb's law, the magnitude of the electrostatic force between the 5.0-nC charge and Q1 can be calculated as follows:

F1 = k * |Q1 * Q3| / r1^2

Substituting the values:

F1 = (8.99 x 10^9 N*m^2/C^2) * |(6.0 x 10^-9 C) * (5.0 x 10^-9 C)| / (0.3 m)^2

Calculate F1 using the given values.

Using Coulomb's law, the magnitude of the electrostatic force between the 5.0-nC charge and Q2 can be calculated as follows:

F2 = k * |Q2 * Q3| / r2^2

Substituting the values:

F2 = (8.99 x 10^9 N*m^2/C^2) * |(-1.0 x 10^-9 C) * (5.0 x 10^-9 C)| / (0.1 m)^2

Calculate F2 using the given values.

Now, to find the net electrostatic force on the 5.0-nC charge due to the other charges, we need to calculate the vector sum of F1 and F2. To do this, we can add the magnitudes of the forces and determine the direction of the net force.

After calculating F1 and F2, add the magnitudes of the forces to find the net electrostatic force.

To determine the direction of the net electrostatic force, we need to consider the signs of the forces. If the forces have the same sign (both positive or negative), they will point in the same direction. If the forces have different signs, they will point in opposite directions.

Therefore, by considering the signs of F1 and F2, we can determine the direction of the net electrostatic force on the 5.0-nC charge due to the other charges.

To find the net electrostatic force on the 5.0-nC charge, we need to calculate the forces exerted on it by each of the other charges and then find the vector sum of these forces.

The electrostatic force between two charges can be calculated using Coulomb's Law, which states that the force (F) between two charges (Q1 and Q2) is given by:

F = k * (Q1 * Q2) / r^2

Here, k is the electrostatic constant (k = 8.99 x 10^9 N•m^2/C^2), Q1 and Q2 are the charges in coulombs, and r is the distance between the charges in meters.

Let's calculate the force exerted by Q1 on the 5.0-nC charge (Q3). The distance between them is given as 0.30 m. Plugging in the values into Coulomb's Law:

F1 = (8.99 x 10^9 N•m^2/C^2) * [(6.0 x 10^-9 C) * (5.0 x 10^-9 C)] / (0.30 m)^2

Calculating this value, we get:

F1 = 5.994 N

Note: The force is positive because Q1 and Q3 are both positive charges.

Next, let's calculate the force exerted by Q2 on Q3. The distance between them is given as 0.10 m. Plugging in the values into Coulomb's Law:

F2 = (8.99 x 10^9 N•m^2/C^2) * [(-1.0 x 10^-9 C) * (5.0 x 10^-9 C)] / (0.10 m)^2

Calculating this value, we get:

F2 = -1797 N

Note: The force is negative because Q2 is a negative charge and Q3 is positive.

To find the net force, we add the forces F1 and F2:

Net force = F1 + F2 = 5.994 N - 1797 N

Net force = -1791 N

To find the magnitude of the net force, we take the absolute value:

Magnitude of net force = |Net force| = |-1791 N| = 1791 N

So, the magnitude of the net electrostatic force on the 5.0-nC charge due to the other charges is 1791 N.

To find the direction of the net force, we consider the signs of the individual forces. Since F2 is negative and F1 is positive, the net force is pointing in the negative direction along the x-axis.

Therefore, the direction of the net electrostatic force on the 5.0-nC charge due to the other charges is in the negative x-direction.