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 = 0.135 m). The charges on the circle are -3.20 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
_________°
To find the magnitude and direction of the net electrostatic force acting on the charge at the center, we can use the principle of superposition, which states that the net electrostatic force on a charge is the vector sum of the individual forces exerted by other charges.
First, let's find the magnitude of the individual forces between the charge at the center and each of the charges on the compass.
1. Force between the charge at the center and the charge at the position due north:
The distance between these two charges is equal to the radius of the compass (0.135 m). Using Coulomb's law, we can calculate the magnitude of the force:
Force1 = (k * |q1| * |q2|) / r^2
= (9.0 x 10^9 Nm^2/C^2 * 3.00 x 10^-6 C * 3.20 x 10^-6 C) / (0.135 m)^2
2. Force between the charge at the center and the charge at the position due east:
The distance between these two charges is also equal to the radius of the compass (0.135 m). Using Coulomb's law, we can calculate the magnitude of the force:
Force2 = (k * |q1| * |q3|) / r^2
= (9.0 x 10^9 Nm^2/C^2 * 3.00 x 10^-6 C * 5.00 x 10^-6 C) / (0.135 m)^2
Now that we have the magnitudes of the individual forces, we can find the net electrostatic force by finding their vector sum.
Net Force = √(Force1^2 + Force2^2)
To find the direction of the net electrostatic force, we can use trigonometry. We can define the direction relative to due east as the angle between the net force vector and the positive x-axis. We can calculate this angle using the inverse tangent function:
Direction = tan^(-1)(Force1 / Force2)
Now you can perform the calculations to find the magnitude and direction of the net electrostatic force acting on the charge at the center.
There are two forces acting on the charge at the center.
Due to the -3.2 uC charge at the north point, there is a repulsion force of
F1 = k*(3.0)*(3.2)*10^-12/(0.135)^2
and it points to the south.
Due to the +5.0 uC charge at the east point, there is an attraction force of
F2 = k*(5.0)(3.2)*10^-12/(0.135)^2
and it points to the east.
k is the Coulomb's Law constant,
9.0 x 10^9 N • m2 / C
Do the multiplication and add the two forces vectorially. The direction of the resultant force will be
arctan 5.0/3.0 = 59.0 degrees south of (clockwise from) east, since 5/3 is the ratio of the two component forces.