Original Question:
Neutral metal sphere A, of mass 0.10kg, hangs from an insulating wire 2.0m long. An identical metal sphere B, with charge -q is brought into contact with sphere A. Sphere A goes 12 degrees away from Sphere B. Calculate the initial charge on Sphere B.
Note: when one object with charge Q is brought in contact with a neutral object 1/2 the charge is transferred to the neutral object.
I don't understand how to do this.
Ans: 3.9x10^-6C
Posted by drwls:
After the spheres touch, each one acquires a charge of -q/2, and the Coulomb repulsion force pushes them away from each other. If each one hangs inclined A = 12 degrees, and T is the tension in the wire,
T cos A = M g
T sin A = k (Q/2)^2/(2L sinA)^2
k is the Coulomb constant, 8.99 x 10^9 N�m^2/C^2
2L sin A is the separation of the spheres
T can be eliminated by dividind one equation by the other
tan A = k (Q/2)^2/(2L sinA)^2/(Mg)
M g tan A = k Q^2/(16 L sin A)^2
This should let you solve for Q
Question - I don't understand what Mg is supposed to be. Also how did you get 16? and on the right side? Is that supposed to be the tension?
I found the distance between A and B to be 0.4158. Can you please explain the rest? I'm not getting the right answer.
Sure! Let's go through the explanation step by step to help you understand how to solve the problem.
First, let's clarify the equation M g tan A = k Q^2/(16 L sin A)^2:
- M stands for the mass of the sphere A.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- A is the angle at which sphere A hangs (which is given as 12 degrees in the problem).
- k is the Coulomb constant, which has a value of 8.99 x 10^9 N·m^2/C^2.
- Q is the charge on sphere B.
- L is the length of the wire, which is given as 2.0 m in the problem.
Now, let's understand how we arrived at the equation.
1. First, we need to recognize that the tension in the wire (T) can be divided into two components: T cos A and T sin A. T cos A supports the weight of the sphere, while T sin A provides the necessary force for the electrostatic repulsion between the spheres.
2. We know that T cos A is equal to M g (where M is the mass of sphere A and g is the acceleration due to gravity). This is because the weight of the sphere is balanced by the vertical component of the tension in the wire.
3. For the horizontal component (T sin A), we can relate it to the electrostatic force using Coulomb's Law. The electrostatic force between the spheres is given by F = k * (Q/2)^2 / (2L sin A)^2, where Q/2 represents the charge acquired by each sphere after they touch.
4. We can equate T sin A to the electrostatic force and solve for Q. This gives us:
T sin A = k * (Q/2)^2 / (2L sin A)^2
5. Now, let's eliminate T from the equation by dividing T sin A (from step 4) by T cos A (from step 2):
tan A = k * (Q/2)^2 / (2L sin A)^2) / (Mg)
6. Simplifying the equation further, we get:
M g tan A = k * Q^2 / (16 L sin A)^2
Here, we obtain 16 on the denominator of the right side by squaring (2L sin A). This arises from the trigonometric identity sin^2 x + cos^2 x = 1.
Now, let's discuss how to continue solving the problem.
You mentioned that you calculated the distance between A and B to be 0.4158, which is the correct separation (2L sin A). From here, you need to solve the equation M g tan A = k Q^2 / (16 L sin A)^2 for Q.
1. Substitute the known values into the equation:
(0.10 kg) * (9.8 m/s^2) * tan(12 degrees) = (8.99 x 10^9 N·m^2/C^2) * Q^2 / (16 * (2.0 m) * (0.4158))^2
2. Rearrange and solve for Q:
Q^2 = [(0.10 kg) * (9.8 m/s^2) * tan(12 degrees) * 16 * (2.0 m) * (0.4158))^2] / (8.99 x 10^9 N·m^2/C^2)
Q^2 = 6.46 x 10^-10 C^2
Q = ± 3.9 x 10^-6 C (taking the positive value since the charge is specified as -q in the problem)
So, the initial charge on sphere B is approximately 3.9 x 10^-6 C.
I hope this explanation helps you understand the problem better and how to arrive at the correct answer. Let me know if you have any further questions!