Three point charges, A = 2.2 µC, B = 7.5 µC, and C = -3.9 µC, are located at the corners of an equilateral triangle . Find the magnitude and direction of the electric field at the position of the 2.2 µC charge.
magnitude = .
N/C
direction =
You need to specify the length of the sides of the triangle, R.
Then compute the two vector forces acting on the 2.2 uC charge. The 7.5 uC charge contributes a repulsive force
F1 = k*2.2*7.5*10^-12/R^2 Newtons
where k is the Coulomb constant and R is the side length in meters.
The other charge (-3.9 uC) will contribute an attractive force, given by the same formula,
F2 = k*3.9*2.2*10^-12/R^2 Newtons,
directed at 120 degrees to F1. Add them as vectors.
To find the magnitude and direction of the electric field at the position of the 2.2 µC charge, we can use the principle of superposition. According to this principle, the total electric field at a point is the vector sum of the electric fields created by each individual charge.
To calculate the electric field at the position of the 2.2 µC charge, we need to find the electric field contributions from charges B and C. Since the triangle is equilateral, the electric field contributions from charges B and C will have the same magnitude but opposite directions.
First, let's calculate the electric field contribution from charge B at the position of the 2.2 µC charge (A). We can use Coulomb's Law to calculate the electric field magnitude:
Electric Field Magnitude (EB) = k * |charge B| / r^2
where k is the electrostatic constant (9 x 10^9 N m^2/C^2), |charge B| is the magnitude of charge B (7.5 µC), and r is the distance between charge B and charge A.
Since the triangle is equilateral, the distance r between charge B and charge A is the length of any side of the triangle. Let's call this length "s". Therefore, r = s.
Electric Field Magnitude (EB) = (9 x 10^9 N m^2/C^2) * (7.5 x 10^-6 C) / s^2
Next, let's calculate the electric field contribution from charge C at the position of the 2.2 µC charge (A). Again, the magnitude of this electric field contribution will be the same as the magnitude of the electric field contribution from charge B, but the direction will be opposite.
Electric Field Magnitude (EC) = k * |charge C| / r^2
where |charge C| is the magnitude of charge C (-3.9 µC). Since charge C is on the opposite side of the triangle, the distance r between charge C and charge A is also the length of any side of the triangle. Therefore, r = s.
Electric Field Magnitude (EC) = (9 x 10^9 N m^2/C^2) * (3.9 x 10^-6 C) / s^2
Finally, we can calculate the total electric field at the position of the 2.2 µC charge (A) by summing the electric field contributions from charges B and C:
Electric Field Magnitude (EA) = EB + EC
To find the direction of the electric field, we need to consider the vector sum of the electric field contributions from charges B and C. Since the magnitudes are the same but the directions are opposite, their vector sum will depend on the relative angles of the triangle.
In an equilateral triangle, the angles between the charges and the point of interest are all 60 degrees. Due to the symmetry of the triangle, the vector sum of the electric field contributions from charges B and C will be directed towards the center of the triangle.
Therefore, the answer is:
Magnitude of Electric Field (EA): The sum of the magnitudes of the electric field contributions from charges B and C.
Direction of Electric Field (EA): Towards the center of the equilateral triangle.
Please note that you will need to substitute the actual values (charge magnitudes and distance) into the formulas provided to calculate the final magnitude and direction of the electric field.