1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Electric field in a cylindrical capacitor

  1. Jul 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A very long question. First let's solve the case for E-field between outer and inner capacitor.


    2. Relevant equations

    3. The attempt at a solution
    Since there is no charge between the 2 capacitors, the charge density is zero.
    By Gauss's law, the integral is zero.

    But I know it doesn't make sense, so what's wrong?


    Attached Files:

    • 1.jpg
      File size:
      48.4 KB
    • 2.jpg
      File size:
      38.7 KB
  2. jcsd
  3. Jul 20, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi athrun200! :smile:

    You cannot draw the conclusion that E is zero from this.

    What you do have is that the surface integral of E is zero according to Gauss's law.
    That means that the flux in of the electric field on one side must equal the flux out of the electric field on the other side.
  4. Jul 20, 2011 #3
    After knowing that there is no net flux flow, how can I get E-feild?
  5. Jul 21, 2011 #4


    User Avatar
    Science Advisor
    Homework Helper

    Are you sure you know what Gauss' law states? It says the flux through the surface is proportional to the enclosed charge, yes? Between the two cylinders there is definitely a nonzero enclosed charge and a nonzero flux. The outer charge is not 'enclosed' if the radius of the surface is between the radii of the two cylinders. You need to choose a cylindrical surface and use the symmetry of the problem to compute the flux.
    Last edited: Jul 21, 2011
  6. Jul 21, 2011 #5

    Do you mean this?

    But I encounter 2 problems.
    1. I don't know the range for z.
    2. Even I don't the range for z, how can I obtain the E from left hand side?

    Attached Files:

    • 1.jpg
      File size:
      54.1 KB
  7. Jul 21, 2011 #6


    User Avatar
    Science Advisor
    Homework Helper

    That is not Gauss' law. Gauss' law relates a surface integral of the E field to a volume integral of the charge rho. Why do you have a volume integral of div(E)? You don't even need to write component integrals to do this problem. E is equal and pointed outward from your surface everywhere using symmetry, right? So the surface integral is just the multiple of the area of the surface per unit length times E. Ditto on the enclosed charge, it's just the amount of charge enclosed per unit length. Notice you are given the charge in terms of coulombs/meter, i.e. charge per unit length. Just equate them. The z range doesn't matter since everything is 'per unit length'.
  8. Jul 21, 2011 #7

    I like Serena

    User Avatar
    Homework Helper

    Take anything for the range of z. Let's call it delta z.

    On your first scan you wrote a surface integral over a closed surface.
    Do you know what it means and what it is for?

    Suppose we know what E is, what would be its flux over the surface of the volume that you just integrated?
  9. Jul 21, 2011 #8
    I thought that it could be easier after applying divergence theorem.


    Is it ok now?

    Attached Files:

    • 1.jpg
      File size:
      32.4 KB
  10. Jul 21, 2011 #9
    I know how to calculate surface, volume integrals.
    And also I know green's and divergence theorem.

    I have done a lot of exercises on them and I can master them very well.

    But when it comes to physical application, I don't know why it is confusing. Maybe I am not good at physics.:uhh:
  11. Jul 21, 2011 #10

    I like Serena

    User Avatar
    Homework Helper

    I guess it takes a bit of effort to wrap the mind around what these mathematical theorems mean and what they are for.
    That's what these exercises are for! ;)

    Much better yes. :)

    So you've calculated an E here.
    But what E is that exactly?
    At which coordinates and in which direction would this be the solution for E?
  12. Jul 21, 2011 #11

    Between [itex]^{R}1[/itex] and [itex]^{R}2[/itex]?
    The direction is normal to the surface?
  13. Jul 21, 2011 #12

    I like Serena

    User Avatar
    Homework Helper

    A little more accurate would be that you solved the radial component of E, which is indeed normal to the surface.

    At which radius did you do the integration?

    Btw, this brings up the question what the other components are.
    Any thoughts on what the component of E in the z direction is?
    And in the theta direction?
  14. Jul 21, 2011 #13
    Do you mean that despite that we consider the case for R1 to R2, the solution of E is form 0 to R1?(Since the range of integral is 0 ro R1)

    It seems E in the z direction is zero, otherwise current flows inside.

    E is constant for all value of theta.
  15. Jul 21, 2011 #14

    I like Serena

    User Avatar
    Homework Helper

    No, you did a "surface" integral for E, not a volume integral (you did that for the charge).

    Basically you selected an cylindrical object with a certain radius and height.
    You integrated E over the surface of this cylinder.
    And you integrated the charge over the volume of this cylinder.

    What was the radius of this cylinder?
    Or rather, at what radius did you calculate E?
    Suppose we would pick another radius r between R1 and R2, what would the radial component of E be then?

    And yes, the z component of E is zero, because otherwise a current would flow in the z direction, which would make no sense since the charge at higher z is the same as the charge at lower z.

    Now if E is constant for all value of theta in the direction of theta, what would happen?
  16. Jul 21, 2011 #15
    Oh, I know what you are talking about.
    The value of E I have obtained is located at R1.
    The value of E decrease when we move from R1 to R2.

    So E is not constant in the direction of theta, it is decreasing.
  17. Jul 21, 2011 #16

    I like Serena

    User Avatar
    Homework Helper

    Indeed the value of E would decrease when we move from R1 to R2, but by how much?
    So E is not constant in the direction of r, it is decreasing.

    However, in the direction of theta things are different.
    Because that is at constant r.

    I get the impression you're mixing up something.
    E is a vector field.
    That means that at any point (r, θ, z) E has 3 components.
    An Er component in the direction of r.
    An Eθ component in the direction of θ.
    And an Ez component in the direction of z.

    I think we've established that Ez is zero.
    And that Er is decreasing when r increases.
    The remaining question is, what is Eθ?
  18. Jul 21, 2011 #17
    Eθ= Er because you have just mentioned that E depends on r.

    For the question of how much Er decrease, I totally have no idea.
    Can I obtain it via diff.?
  19. Jul 21, 2011 #18

    I like Serena

    User Avatar
    Homework Helper

    No, this is not right.
    Basically, the only thing I said, is that Er depends on r.

    Actually you have 3 functions here that you need to solve:
    Er(r, θ, z)
    Eθ(r, θ, z)
    Ez(r, θ, z)

    The component Er(r, θ, z) it the one (the *only* one) that is perpendicular to the surface of the cylinder that you integrated.
    This component will be the same for any θ and for any z.
    But it will change if r changes.
    We typically indicate this dependency by writing Er = Er(r).

    Can you say something about the dependencies of Ez(r, θ, z)?
    Is it dependent of either r, θ, or z?
    And Eθ(r, θ, z)?

    This time round you integrated over a cylinder with radius R1.
    Can you do the same integration over a cylinder with radius r?
  20. Jul 21, 2011 #19
    It seems Ez and Eθ are independent of the three variables.
    The value of them are 0 too.
    If it is the case, I have only 1 equation that is Er.

    I am still very puzzle on this question. Can you do the case of R1<r<R2 for me?

    I would like to have an example as a reference and try to do the rest of the cases.
  21. Jul 21, 2011 #20

    I like Serena

    User Avatar
    Homework Helper

    Ez = Eθ = 0

    As doing the case for you - it's exactly what you did, except with r instead of R1 on the left hand side where you have the surface integral of E.
    This means you integrate over a cylinder that is bigger than the inner cylinder.
    The contained charge is the same (the right hand side), but the surface integral changes.
    Last edited: Jul 21, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook