Hanging pith ball problem with unequal masses and charges

In summary, two balls with different masses and charges are suspended by identical threads of length 0.10m, making a 25 degree angle at equilibrium with the vertical. The charge of one ball is 2.0 x 10^-7 C and the other is 6.0 x 10^-8 C. Using Coulomb's Law and the equations for weight and tension, the masses of both balls are calculated to be 3.3 x 10^-3 kg. However, the correct answer given in the book is 1.2 x 10^-4 kg, leading to confusion about the relationship between the masses and the equal angles formed by the threads. Further insight is needed to solve this problem.
  • #1
furth721
6
0

Homework Statement



Suppose that two balls are suspended by identical threads of length .10m anchored to the same point. The pith balls have different masses and charges. The charge of one is 2.0 X 10^-7 C and the other one is 6.0 X 10^-8 C. Both threads make a 25 degree angle at equilibrium with the vertical. What is the mass of each ball?



Homework Equations



culombs law = kq1q2/r^2

weight of ball 1 = m1g

weight of ball 2 = m2g

Tension in x direction = Tsin(25) = kq1q2/r^2

Tension in y direction of ball 1 = Tcos(25) = m1g

Tension in y direction of ball 2 = Tcos(25) = m2g

r = distance between charges = .10sin(25)

The Attempt at a Solution



I drew a free body diagram of the first ball and set the x and y components of the tension equal to the weight and the force of the like charges applied on ball one. Since both strings make a 25 degree angle with the vertical the pith balls would be the same distance away from the vertical and have equal tensions and masses. This is where I think I am making a mistake. I solve for the mass of the first ball by solving for T in one equation and subbing that into the other equation. I get that the mass of both balls are something like 3.3 x 10 ^ -3 kg. The answer in the back of the book says that the masses of both balls is 1.2 x 10 ^ -4 kg. I would really appreciate if someone could give me their input on the problem.
 
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  • #2
The force due to gravity (its weight), the force due to the tension and the electrostatic repulsive force should all cancel out to give a net force of zero since the balls are no longer moving.

Also, the distance between charges is incorrect I think, that's half the distance. It should be (0.10*sin (25))*2
 
  • #3
Oh yeah, the distance is .20sin(25) that was a typo on my part. I understand that because I've done problems like this before where the masses and charges were equal. To solve it you have to set the weight equal to Tcos(25) which is the y component of the tension and the force of the charge equal of the x component then solve for T and substitute into the other equation. The thing I am having trouble with is, the balls are supposed to be two different masses but they make the same angle with the vertical which means that the tension acting on both of the balls is equal, and the Force of charge on the balls is equal shouldn't that make the mass equal as well?
 
  • #4
Hmm, that's interesting. I actually didn't notice it stated different masses until you said it now. I would've thought that being two different masses the angles should not be the same provided all other things remain unchanged for both balls.

I'm stumped, maybe someone else can provide some insight into this, because I'm interested to know as well.
 
  • #5


I would suggest approaching this problem by breaking it down into smaller parts and using Coulomb's Law to calculate the tension in the strings. First, calculate the distance between the two charges (r) by using the given length of the strings and the angle they make with the vertical. Then, use Coulomb's Law to calculate the tension in the strings (T) by setting the force of repulsion between the two charges equal to the tension in the strings. Once you have the value of T, you can use it to calculate the mass of each ball by equating the tension in the strings to the weight of the balls. This approach should give you the correct answer of 1.2 x 10^-4 kg for the mass of each ball. It is important to carefully set up and solve each equation, making sure to use the correct values and units. If you are still having trouble, I would suggest seeking help from a teacher or classmate.
 

Related to Hanging pith ball problem with unequal masses and charges

1. What is the hanging pith ball problem with unequal masses and charges?

The hanging pith ball problem with unequal masses and charges is a physics problem that involves two charged and unequal masses hanging from a string and interacting with each other due to their charges.

2. What is the significance of this problem?

This problem is significant because it helps to demonstrate the principles of electrostatics, specifically the concept of electrostatic force, and how it affects objects with different charges and masses.

3. How is this problem solved?

This problem can be solved using the principles of Coulomb's Law, which states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

4. What factors affect the behavior of the masses in this problem?

The behavior of the masses in this problem is affected by their charges, masses, and the distance between them. The greater the difference in charges and masses, and the closer they are to each other, the stronger the electrostatic force between them will be.

5. How does this problem relate to real-life situations?

This problem can be used to explain the behavior of objects with different charges, such as ions in a plasma or particles in a cloud. It also has practical applications in areas such as electrostatics and electromagnetism, which are important in modern technology.

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