Advanced Problem in Electrostatics

In summary, the question is asking for the horizontal velocity needed for a charged ball on a string to rotate in a vertical plane in a uniform electrostatic field. The tension in the string in the lower position must be 15 times the weight of the ball. The solution involves equating gravitational, electric, and centrifugal forces and includes a diagram to aid understanding.
  • #1
zorro
1,384
0

Homework Statement


A ball of mass m with a charge q can rotate in a vertical plane at the end of a string of length l in a uniform electrostatic field whose lines of force are directed upwards. What horizontal velocity must be imparted to the ball in the upper position so that the tension in the string in the lower position of the ball is 15 times the weight of the ball.


The Attempt at a Solution



I don't get what is the meaning of the question. How can a sphere rotate on a string as its tangent? Why will tension in the string change if ball is given a velocity? When it is imparted with a velocity, does the ball rotate purely or translates too?

Don't explain how to solve, I just need help in understanding the question.
Perhaps a figure might help.
Thanks
 
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  • #2
It's hard to know what the question is asking, I think. Was a diagram included?

Perhaps the ball is on the end of the string with the string's other end fixed in space. Then the ball would be confined to a disk around that point. Also, by "rotation," could it mean that the ball is moving in a circle around that point rather than rotating on its own axis?

If that is the case then it would make sense that the tension on the string depends on the ball's velocity because of centrifugal force.
 
  • #3
Imagine you have one end of the string in your hand, at the other end is the charged sphere. Now you can slowly start swinging the sphere like a pendulum. If you go faster and faster, at some point the sphere will no longer swing back, but complete the circle. It will rotate in a vertical plane.
Now you just have to equate gravitational, electric and centrifugal force to find the answer.

Btw. crazy teacher who defined this problem to be electrostatics. There is not much static going on.
 
  • #4
betel said:
Imagine you have one end of the string in your hand, at the other end is the charged sphere. Now you can slowly start swinging the sphere like a pendulum. If you go faster and faster, at some point the sphere will no longer swing back, but complete the circle. It will rotate in a vertical plane.
Now you just have to equate gravitational, electric and centrifugal force to find the answer.

Oh! I got it.
I thought the sphere rotates about its own axis on the string.
Thanks
 
  • #5
.

I can provide some clarification on the question and offer a potential solution approach. The problem is asking for the required horizontal velocity to be imparted to the ball in order for the tension in the string to be 15 times the weight of the ball. This means that the ball will experience a greater downward force due to the tension in the string, resulting in a higher tension force to counteract the weight of the ball.

To solve this problem, we can use the principles of electrostatics, specifically Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. In this case, the ball has a charge q and is in a uniform electrostatic field, so we can use the formula F = qE, where E is the electric field strength.

In the upper position, the ball is stationary and the tension in the string is equal to the weight of the ball, mg. As the ball rotates downward, the tension in the string increases due to the electric field, and at the lower position, it is 15 times the weight of the ball. This means that the electric force on the ball is 15mg.

To find the required horizontal velocity, we can use the principle of conservation of energy. At the upper position, the ball has potential energy due to its height, and at the lower position, it has both potential and kinetic energy. We can equate these energies to find the velocity.

mg(2l) = 15mg(l) + 1/2mv^2

Simplifying, we get v = √(3gl). This is the required horizontal velocity to impart to the ball at the upper position in order for the tension in the string at the lower position to be 15 times the weight of the ball.

I hope this helps in understanding the problem and provides a potential solution approach. A figure or diagram would indeed be helpful in visualizing the scenario.
 

1. What is the difference between electrostatics and electrodynamics?

Electrostatics is the study of stationary electric charges and the forces between them, while electrodynamics is the study of how electric charges behave in motion and how they interact with magnetic fields.

2. What is the equation for Coulomb's Law?

Coulomb's Law states that the force between two point charges is directly proportional to the product of the charges 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 Coulomb's constant, q1 and q2 are the charges, and r is the distance between the charges.

3. How do you calculate the electric field due to a point charge?

The electric field due to a point charge can be calculated using the equation E = k * (q/r^2), where E is the electric field, k is the Coulomb's constant, q is the charge, and r is the distance from the point charge.

4. What is the principle of superposition in electrostatics?

The principle of superposition states that the total electric field at any point due to a collection of point charges is equal to the vector sum of the individual electric fields at that point. In other words, the electric field at a point is the sum of the electric fields from each individual charge present at that point.

5. What is the significance of Gauss's Law in electrostatics?

Gauss's Law is a fundamental law in electrostatics that relates the electric flux through a closed surface to the charge enclosed by that surface. It allows us to easily calculate the electric field for highly symmetric charge distributions and is also useful in determining the net charge contained within a closed surface.

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