1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Potential at centre of cylinder

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=59298&stc=1&d=1370417090.jpg


    2. Relevant equations



    3. The attempt at a solution
    I have no idea on this one. I can't understand what happens when some portion of the cylinder is earthed and other is maintained at 110V. The potential of the earthed portion is 0V but how does that even help me? :confused:

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Jun 5, 2013 #2
    Is this not just a simple application of the superposition of electric potentials?
     
  4. Jun 5, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Don't you need to know the charge distributions to do that? The OP is a Dirichlet boundary value problem, no?
     
  5. Jun 5, 2013 #4
    What is this Dirichlet boundary value problem? :confused:
     
  6. Jun 5, 2013 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Suppose you choose an imaginary circle of radius r < R lying in a plane perpendicular to the axis of the cylinder and centered on the axis of the cylinder.

    Try to show that the integral ∫0 V(r,θ)dθ is independent of the value of r, where the integral is over the circle of radius r and θ is the polar angle.

    If you can do that, then you can relate V on the axis to the values of V on the cylinder.
     
    Last edited: Jun 5, 2013
  7. Jun 5, 2013 #6
    I understand what you say here but how am I even supposed to find ##V(r,\theta)##? :confused:

    I don't even have the charge distribution, only the potential of two portions of cylinder.
     
  8. Jun 5, 2013 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You don't need to find V(r, θ) in order to show that ∫V(r,θ)dθ is independent of r.

    Although not obvious, you can show this by starting with Gauss's law for a closed cylindrical Gaussian surface of radius r as shown. See if you can express the net flux through the surface in terms of an integral of Er, the radial component of the electric field. Then relate Er to V(r,θ) inside the integral.
     

    Attached Files:

  9. Jun 5, 2013 #8
    Flux through the Gaussian surface is ##\phi=E_r\cdot 2\pi rh## but this does not involve any integral. :confused:

    I think I cannot consider ##E_r## as constant over the whole Gaussian surface, right?

    From Gauss law, ##\phi=\int \vec{E_r}\cdot \vec{dA}##. ##\displaystyle E_r=-\frac{dV}{dr}## but this does not involve ##\theta##. :confused:
     
  10. Jun 5, 2013 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Er cannot be assumed to be constant over the surface. So, can you bring it out of the flux integral?

    [EDIT]
    Since V is a function of r and θ, ∂V/∂r will also be a function of r and θ.
     
    Last edited: Jun 5, 2013
  11. Jun 5, 2013 #10
    I don't know how to solve ##dV(r,\theta)/dr##. :confused:
     
  12. Jun 5, 2013 #11

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You don't need to solve dV/dr. Express the flux integral in terms of ∂V/∂r. The goal is to use the flux integral to show d/dr ∫V(r,θ)dθ = 0, which means ∫V(r,θ)dθ over a circle of radius r is a constant (independent of the size of the circle).
     
    Last edited: Jun 5, 2013
  13. Jun 5, 2013 #12

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    See http://physics.usask.ca/~hirose/p812/notes/Ch2.pdf.
    If the potential is specified on a closed surface, the potential off the surface is uniquely determined in terms of the surface potential. This is known as Dirichlet's boundary value problem
     
  14. Jun 6, 2013 #13
    Do I need to simply substitute ##E_r=∂V/∂r##? :confused:

    That's too much mathematics. Is there no easier way to this problem? :confused:
     
  15. Jun 6, 2013 #14
    Well, the solution to a Dirichlet problem, involving Green's function, is an application of superposition :)

    But you are correct in that it is not simple and not direct as my original suggestion implied. I think I misunderstood the problem initially.
     
    Last edited: Jun 6, 2013
  16. Jun 6, 2013 #15
    This is the standard approach to such problems. Besides, most of mathematics there is not for you do, but merely to follow. If you follow, you will see in one example that most of what needs to be done for this problem is done there, you just need to plug in your data to get the solution.
     
  17. Jun 6, 2013 #16

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. What does Gauss' law look like for the chosen Gaussian surface after making the substitution ##E_r=-∂V/∂r##.

    That should bring you to within a few simple steps of the answer to the question.
     
  18. Jun 6, 2013 #17
    ##\displaystyle \phi=\int -\frac{∂V}{∂r}dA##

    But what is ##dA##? :confused:
     
  19. Jun 6, 2013 #18

    TSny

    User Avatar
    Homework Helper
    Gold Member

    What part of the Gaussian surface are you integrating over here?
     
  20. Jun 6, 2013 #19
    Is ##dA=rd\theta dy## where y is the distance of the patch(dA) from the top base of the Gaussian surface? :confused:
     

    Attached Files:

  21. Jun 6, 2013 #20

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Potential at centre of cylinder
Loading...