Uniformly Charged Ring Acting on a Particle

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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 is derived using the formula E=kQz/(z^2+a^2)^(3/2), leading to the force F=kqQz/(z^2+a^2)^(3/2). To determine potential energy, the work done by the electric force is expressed as W=∫F·dl, with the potential energy difference calculated using U(B)-U(A) = -∫(F·dl). The zero point for potential energy is conventionally set at infinity in electrostatics.

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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.
 
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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:

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

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
 

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