A charge of -3.00 µC is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass (radius = 0.125 m). The charges on the circle are -4.60 µC at the position due north and +5.00 µC at the position due east. What is the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east (0°).
Magnitude
_____N
Direction
______°
Magnitude:
(8.99e9)(4.6e-6)(3.0e-6)/ (0.125)^2
=7.94 N
(8.99e9)(5.0e-6)(3.0e-6)/ (0.125)^2
=8.63 N
Adding both values as vectors:
sqrt[(7.94N squared)+(8.63N squared)]
Magnitude= 11.73 N
Direction:
theta = inverse tan (7.94/8.63)
= 42.62 degrees South of East
LOOks good to me!
To find the magnitude of the net electrostatic force acting on the charge at the center of the compass, we can use Coulomb's law. Coulomb's law states that the force between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. The equation for Coulomb's law is:
F = (k * |q1| * |q2|) / r^2
where F is the force, k is the electrostatic constant (8.99e9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
In this case, we have three charges - the charge at the center and the two charges on the circle. The charge at the center has a magnitude of -3.00 µC (-3.0e-6 C). The charge on the circle due north has a magnitude of -4.60 µC (-4.6e-6 C), and the charge on the circle due east has a magnitude of +5.00 µC (5.0e-6 C).
The distance between the charges is the radius of the circle, which is given as 0.125 m.
To find the magnitude of the net electrostatic force on the charge at the center, we can calculate the forces between the center charge and each of the charges on the circle using Coulomb's law, and then add them together as vectors.
The force between the center charge and the charge on the circle due north is given by:
F1 = (k * |q1| * |q2|) / r^2
= (8.99e9 N m^2/C^2) * (3.0e-6 C) * (4.6e-6 C) / (0.125 m)^2
= 7.94 N
The force between the center charge and the charge on the circle due east is given by:
F2 = (k * |q1| * |q2|) / r^2
= (8.99e9 N m^2/C^2) * (3.0e-6 C) * (5.0e-6 C) / (0.125 m)^2
= 8.63 N
To find the net force, we can add the vectors F1 and F2 using vector addition. The magnitude of the net force is given by:
F = sqrt(F1^2 + F2^2)
= sqrt((7.94 N)^2 + (8.63 N)^2)
= 11.73 N
So, the magnitude of the net electrostatic force acting on the charge at the center of the compass is 11.73 N.
To find the direction of the net force, we can use trigonometry. The direction is specified relative to due east (0°).
tan(θ) = (F1 / F2)
θ = inverse tan(F1 / F2)
= inverse tan(7.94 N / 8.63 N)
= 42.62°
Therefore, the direction of the net electrostatic force is 42.62 degrees south of east.