A charge of -6.8 micro-Coulombs is placed at x = 0, y = 0. A charge of -18.5 micro-Coulombs is placed at x = 0, y = +37 cm. Another charge of -12.9 micro-Coulombs is placed at x = +17 cm, y = 0 cm. What is the angle the total electric force on the charge at x = +17 cm, y = 0 cm makes when measured from the positive x direction, counter-clockwise? Answer in degrees
To find the angle the total electric force on the charge at x = +17 cm, y = 0 cm makes when measured from the positive x direction, counter-clockwise, we can use the concept of vector addition.
The electric force between charges is a vector quantity, meaning it has both magnitude and direction. It follows the principle of superposition, which states that the total force on a charge due to multiple other charges is the vector sum of the individual forces exerted by each charge.
Let's break down the problem into steps:
Step 1: Find the magnitude and direction of the force between charges (-6.8 μC) and (-18.5 μC).
Using Coulomb's Law, the magnitude of the force between two charges is given by:
F = k * |q1 * q2| / (r^2)
where F is the magnitude of the force, k is the electrostatic constant (9 × 10^9 N*m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
The distance between the charges at x = 0, y = 0 cm and x = 0, y = +37 cm is 37 cm (or 0.37 m).
Substituting the given values:
F1 = (9 × 10^9 N*m^2/C^2) * |-6.8 × 10^-6 C * -18.5 × 10^-6 C| / (0.37 m)^2
Calculating F1 will give us the magnitude of the force between -6.8 μC and -18.5 μC.
Step 2: Find the magnitude and direction of the force between charges (-6.8 μC) and (-12.9 μC).
Using the same formula as above and substituting q1 = -6.8 μC, q2 = -12.9 μC, and r = 17 cm (or 0.17 m), we can calculate F2.
Step 3: Find the net force on the charge at x = +17 cm, y = 0 cm by adding the individual forces.
The net force can be obtained by vector addition. Since the forces between charges are vectors, we need to consider both magnitude and direction.
Let F1 and F2 represent the magnitudes of the forces calculated in steps 1 and 2, respectively. Let θ1 represent the angle between the positive x-direction and F1, and let θ2 represent the angle between the positive x-direction and F2. Note that we can find the directions of the forces because they are both repulsive (negative charges repel each other).
Now, using vector addition, the net force Fnet can be found by the following equation:
Fnet = sqrt(F1^2 + F2^2 + 2 * F1 * F2 * cos(180° - θ1 + θ2))
Step 4: Calculate the angle from the positive x-direction.
The angle we want to find is the angle between the net force (Fnet) calculated in step 3 and the positive x-direction. We can calculate it using the equation:
θnet = arccos((F1^2 + Fnet^2 - F2^2) / (2 * F1 * Fnet))
Finally, the answer will be the angle θnet expressed in degrees.
Please note that in this explanation, I assumed that the given charges have fixed positions and the charges don't move. The calculations are based on the assumed static charges.