• Support PF! Buy your school textbooks, materials and every day products Here!

A ring of charge

  • Thread starter Sierra
  • Start date
  • #1
14
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.
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
(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:
  • #3
14
0
(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?
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
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
 
  • #5
14
0
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)
 
  • #6
Simon Bridge
Science Advisor
Homework Helper
17,847
1,644
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.
 

Related Threads on A ring of charge

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
9
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
14
Views
10K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
2
Views
16K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
2
Views
9K
Top