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A ring of charge

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    λ=Q/l
    dl=RdΘ
    dq=λdl
    E(z)=KQx/(x^2+R^2)^3/2
    U= KQq/r
    3. 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.
     
  2. jcsd
  3. Jan 31, 2015 #2

    Simon Bridge

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    (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: Jan 31, 2015
  4. Jan 31, 2015 #3
    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?
     
  5. Feb 1, 2015 #4

    Simon Bridge

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    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
     
  6. Feb 3, 2015 #5
    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)
     
  7. Feb 3, 2015 #6

    Simon Bridge

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    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.
     
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