Uniformly Charged Ring Acting on a Particle

Click For Summary
The discussion focuses on calculating the electric force exerted on a particle located a distance z above a uniformly charged ring with charge Q. The electric field E was derived as E=kQz/(z^2+a^2)^(3/2), leading to the force F=kqQz/(z^2+a^2)^(3/2). To determine the potential energy, the concept of work is applied, using the integral of the force along a chosen path. It is emphasized that potential energy is defined for conservative forces, and the zero point of potential energy is typically set at infinity in electrostatics. The integration path for calculating potential energy can conveniently be along the z-axis.
acedeno
Messages
34
Reaction score
4

Homework Statement


Solve for the Electric force exerted on a Particle a distance z above a uniform ring of charge Q.

Determine the potential energy of the charge where the charge lies directly in the center.

Homework Equations


F=kq1q1/r^2


The Attempt at a Solution


Knowing E=F/q I just solved for E then multiplied it by the charge q.

resulting in E=kQz/(z^2+a^2)^3/2 thus, F=kqQz/(z^2+a^2)^3/2

Now, I'm not sure about how to go about expressing the potential energy.
My intuition tells me to do Work=Potential Energy, so,
W= integral of[F(dot)dl]

I'm not sure where to go from here because I'm not sure as to what I should make dl.
 
Physics news on Phys.org
Potential energy is defined for conservative forces, so as the force is negative gradient of the potential energy. The work of a conservative force when a body moves from point A to B is independent on the path.
You can calculate the potential energy difference by integrating the force along any path form A to B:

U(B)-U(A) =-\int_{A}^{B}(\vec {F} \cdot \vec{dl} )

The zero point of the potential energy is arbitrary. In Electrostatics, it is at infinity in most cases. For the path, you can chose the most convenient one. For your problem, it can be along the z axis.

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
3
Views
985
  • · Replies 5 ·
Replies
5
Views
4K
Replies
4
Views
3K
Replies
4
Views
1K