Electric Potential & Electric Field of a Ring of Charge

  • Thread starter Thread starter Sierra
  • Start date Start date
  • Tags Tags
    Charge Ring
Click For Summary

Homework Help Overview

The problem involves a ring of charge located in the x-y plane, with a specific charge and radius. Participants are tasked with finding the electric potential and electric field at a certain distance above the origin, as well as the speed of a proton released from rest at that height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between electric potential and electric field, with some questioning the correctness of the equations used. There are attempts to clarify the approach to solving part (a) before addressing parts (b) and (c). Others suggest revisiting the electric potential equation to ensure accuracy.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the relationships between the concepts involved. Some guidance has been offered regarding the use of equations, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants express uncertainty about the equations to use and the relationships between electric potential and electric field. There is mention of needing to refer to notes for clarification on these concepts.

Sierra
Messages
14
Reaction score
0

Homework Statement


A ring of charge is situated in the x‐y plane centered about the origin. The ring has a
uniformly distributed charge Q = ‐10 nC and a radius R = 2.0 cm.
a. Find the electric potential at a distance z = 5.0 cm above the origin on the z=axis.
b. Find the electric field at a distance z = 5.0 cm above the origin on the z=axis.
c. Find the speed of a proton as it passes through the origin assuming that it is
released from rest at z=5.0 cm.

Homework Equations


λ=Q/l
dl=RdΘ
dq=λdl
E(z)=KQx/(x^2+R^2)^3/2
U= KQq/r

The Attempt at a Solution



After solving to get this

b) E(z)=KQx/(x^2+R^2)^3/2

=(8.998x10^9)(10x10^-6)(0.05m) / (0.05^2+0.02^2)^3/2
=2.88x10^7 N/C

Now I feel like this is really easy to get a, but I can't seem to get it. And c is just blew my mind.
 
Physics news on Phys.org
(b) and (c) follow most easily from doing (a) first.

So you did (b) first... what is the relationship between the electric field and the electric potential?
Did you mistype the equation for E(z) - you have it dependent on x alone.
... in fact: much much easier for this geometry to just go back and do (a) from scratch.
What is the electric potential at z due to one charge element?

For (c) just use conservation of energy...
 
Last edited:
Simon Bridge said:
(b) and (c) follow most easily from doing (a) first.

So you did (b) first... what is the relationship between the electric field and the electric potential?
Did you mistype the equation for E(z) - you have it dependent on x alone.
... in fact: much much easier for this geometry to just go back and do (a) from scratch.
What is the electric potential at z due to one charge element?

For (c) just use conservation of energy...

So I would use the electric potential equation to get a correct? Just fill what is known? So U=(8.998x10^9)(10x10^-6)(0.05) / (0.02)? Or since you asked what is the relationship. Would it be equal of each other?
 
So I would use the electric potential equation to get a correct?
Since (a) asks you to find the electric potential, then the electric potential equation is what you use.
Just fill what is known? So U=(8.998x10^9)(10x10^-6)(0.05) / (0.02)?
How do you know to use that specific equation?
Or since you asked what is the relationship. Would it be equal of each other?
... would what be equal to each other? Don't guess - go look in your notes.

Guessing that you mean "is the electric field equal to the electric potential?" The answer is "no".

The electric field is the negative gradient of the electric potential.
The problem set is testing your knowledge of this so you should have that in your notes ... go look.
Or see: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/efromv.html
 
Simon Bridge said:
Since (a) asks you to find the electric potential, then the electric potential equation is what you use. How do you know to use that specific equation? ... would what be equal to each other? Don't guess - go look in your notes.

Guessing that you mean "is the electric field equal to the electric potential?" The answer is "no".

The electric field is the negative gradient of the electric potential.
The problem set is testing your knowledge of this so you should have that in your notes ... go look.
Or see: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/efromv.html

I redid a and I got 1.8x10^3 V from ΔV= Kq/r from which it is (8.998x10^9)(10*10^-9)/ (0.05m)
 
I cannot help you if you don't answer questions.
If you are confident in your new answer, you can do the rest of the problem.
 

Similar threads

Replies
11
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
4
Views
4K
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
20
Views
4K