A particle carrying a charge of 6.0uC is located at the origin of a rectangle coordinate system, and a particle carrying a charge of 4.0uC is located at (0,5.0m). What are the magnitude and direction of the electric field at the positions (5.0m,0), (-5.0m,0), and (0,-5.0m)?

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To find the magnitude and direction of the electric field at a given position due to the charges, we can use Coulomb's Law and the principle of superposition.

Coulomb's Law states that the electric field created by a point charge is given by:

E = k * q / r^2

Where:
- E is the electric field strength
- k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)
- q is the charge of the particle creating the electric field
- r is the distance between the point charge and the position where we want to find the electric field

For the position (5.0m,0):

The electric field created by the particle at the origin (charge = 6.0 μC) can be calculated using Coulomb's Law:

E1 = (8.99 x 10^9 N m^2/C^2) * (6.0 x 10^-6 C) / (5.0 m)^2

Simplifying the calculation:

E1 = 215.76 N/C (approximately)

The direction of the electric field at position (5.0m,0) will be towards the positive x-axis because the positive charge at the origin repels positive charges.

For the position (-5.0m,0):

The electric field created by the particle at the origin (charge = 6.0 μC) can be calculated using Coulomb's Law, which will have the same magnitude as E1:

E2 = E1 = 215.76 N/C (approximately)

The direction of the electric field at position (-5.0m,0) will be towards the negative x-axis because the positive charge at the origin repels positive charges.

For the position (0,-5.0m):

The electric field created by the particle at (0,5.0m) (charge = 4.0 μC) can be calculated using Coulomb's Law:

E3 = (8.99 x 10^9 N m^2/C^2) * (4.0 x 10^-6 C) / (5.0 m)^2

Simplifying the calculation:

E3 = 143.84 N/C (approximately)

The direction of the electric field at position (0,-5.0m) will be towards the negative y-axis because the positive charge at (0,5.0m) repels positive charges.

To summarize:

At the position (5.0m,0): The magnitude of the electric field is approximately 215.76 N/C, and the direction is towards the positive x-axis.

At the position (-5.0m,0): The magnitude of the electric field is approximately 215.76 N/C, and the direction is towards the negative x-axis.

At the position (0,-5.0m): The magnitude of the electric field is approximately 143.84 N/C, and the direction is towards the negative y-axis.

To find the magnitude and direction of the electric field at different positions, we can use the principles of Coulomb's law and the concept of superposition. Coulomb's law states that the electric field created by a point charge is directly proportional to the charge and inversely proportional to the square of the distance. The formula for the electric field at a point due to a point charge can be calculated using the following equation:

E = k * (q / r^2)

where E is the electric field, k is the Coulomb's constant (9.0 x 10^9 Nm^2/C^2), q is the charge of the particle, and r is the distance between the point charge and the location where the electric field is being calculated.

Now, let's calculate the electric field at the given positions.

1. Electric field at position (5.0m,0):
To find the electric field at this position, we need to consider the electric fields due to both charges. Let's call the charge at the origin Charge Q1 and the charge at (0,5.0m) as Charge Q2.

Electric field due to Q1:
E1 = k * (Q1 / r^2)
Since Q1 = 6.0 x 10^-6 C and r = 5.0m (distance from origin to (5.0m,0)):
E1 = (9.0 x 10^9 Nm^2/C^2) * (6.0 x 10^-6 C) / (5.0m)^2

Electric field due to Q2:
E2 = k * (Q2 / r^2)
Since Q2 = 4.0 x 10^-6 C and r = 5.0m (distance from (0,5.0m) to (5.0m,0)):
E2 = (9.0 x 10^9 Nm^2/C^2) * (4.0 x 10^-6 C) / (5.0m)^2

To find the total electric field at position (5.0m,0), we need to calculate the vector sum of E1 and E2.

Magnitude of the electric field at (5.0m,0):
E_total = sqrt((E1)^2 + (E2)^2)

Direction of the electric field at (5.0m,0):
The direction can be determined using the angle between the vector sum and the positive x-axis.

2. Electric field at position (-5.0m,0):
The process is similar to finding the electric field at (5.0m,0), but we'll use the distances and charges accordingly.

3. Electric field at position (0,-5.0m):
Again, the process is similar, but this time we'll use the distances and charges accordingly.

By following these steps, you can calculate the magnitude and direction of the electric field at the given positions.

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