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Electric Potential of a hollow cylinder

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A hollow cylinder of radius r and height h has a total charge q uniformly distributed over its surface. The axis of the cylinder coincides with the z axis, and the cylinder is centered at the origin, as shown in the figure.

    What is the electric potential V at the origin?

    2. Relevant equations


    3. The attempt at a solution

    So, essentially, I'm attempting to derive the equation for electric potential at the origin of a cylinder. The problem first asks me to consider the cylinder as a series of rings and find the potential for a given ring. Once I have that, I should be able to integrate that expression to find the potential for the cylinder itself, but I'm having trouble even deriving the potential for a ring.

    We are told that the charge of a ring is dq=qdz/h. Trying to understand this precisely, this means that the charge of a ring is equal to the charge at a given thickness along h? Please feel free to correct me if I'm wrong!

    So given that that is our charge, we can now attempt to find the potential at the origin for a ring, and this is where I can't seem to derive the correct equation.

    dV(z)=(1/4piε)∫dq/r = (1/4piε)∫qdzr/h = (1/4piε)qr/h∫dz

    For this portion, it requests that the answer be in terms of dz, z, q, h, r, and epsilon_0, and I'm pretty sure I've already gone wrong somewhere, in terms of arriving at the solution.

    If someone could please help me out, I'd really appreciate it.

    Thank you!
  2. jcsd
  3. Feb 28, 2012 #2


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    Staff: Mentor

    If the entire cylinder has charge q and the cylinder is of height h, then the charge density per unit of height is q/h. That means some ring of height dz will have total charge qr = (q/z)dz . Essentially that's a horizontal slice of the cylinder of thickness dz. (Personally, I'd let the total cylinder charge be Q and the charge of a ring be q, just to avoid subscripts!)
    You're nearly there. Consider that the ring is floating above the x-y plane at some height z. What's the distance from a point on that ring to the origin? Does it vary as you go around the ring?

    Now consider that the ring has a charge distribution all around its circumference. The radius of the ring is r (same as the cylinder), where r is measured from the z-axis. If that r sweeps around the circumference of the ring then it sweeps over all the charge there, so a differential element of charge on the circumference of the ring is given by

    ## dq_r = \frac{q_r}{2 \pi} d\theta ##

    Each of these dqr's is at a given distance from the origin. You want to sum their potential contributions by integrating over θ. This shouldn't be to tricky because everything else in the expression should be constants for a given ring :wink:

    You should end up with an expression for the potential at the origin due to an individual ring at height z. Your next task will then be to add up the contributions of all the rings that make up the cylinder...
  4. Feb 28, 2012 #3
    Thanks for the response!

    Let's see. The distance from a point on the ring to a point along the axis at some height z would be (r^2 + z^2)^1/2

    Now for dq=q/2*pi*d(theta), I'm not quite understanding. Essentially, dq is the charge for a specific piece of the ring, right? So, the d(theta) makes sense, because it refers to how much circumference is being swept out. I don't understand the 2*pi, part, simply because I would have thought to use 2*pi*r, so I'm a little unclear about that bit.
  5. Feb 28, 2012 #4


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    Staff: Mentor

    It'll probably be easier in the long run if you set the ring at height z and keep the point at the origin. After all, you're eventually going to be summing the effects at the origin from rings up and down the z-axis.
    There's a charge q spread around ##2 \pi## radians of angle. So a bit of angle dθ represents a fraction dθ/##2 \pi## of that charge.

    If you insist on using distances, then the bit of circumference is r dθ and the whole circumference is ##2 \pi r##, hence ##(r d \theta)/(2 \pi r)## . The r's cancel anyways :wink:
  6. Feb 28, 2012 #5
    Ok, so your dq makes much more sense to me than the dq given, which is dq=qdz/h. It seems to me that they're using height of ring which doesn't make any sense to me, because that doesn't at all refer to sweeping out circumferences as the dq you used does.
  7. Feb 28, 2012 #6


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    Staff: Mentor

    Ah, but your ring is just a slice of the whole cylinder. The cylinder has height h, and it has some total charge Q (I'll call it Q to distinguish it from a ring's total charge q). So a charge Q is spread evenly over height h of the cylinder. A slice of thickness dz will thus have charge

    ## dQ = \frac{dz}{h} Q ##

    What you've been working with so far is some arbitrary total charge q on a ring, and finding the potential at the orgin due to that charge on a ring located at some height z. Once you have the expression for that potential you will make the leap of saying that the charge q on the ring is actually dQ, and there's a stack of them making up a cylinder of total charge Q. You should be able to plug in the expression for dQ and set up an integral along the z-axis.
  8. Feb 28, 2012 #7
    Got it! So, integrating the equation for potential from a ring, I get:

    q/4*pi*ε ∫dz/h*((z^2+r^2)^1/2)

    But now the integration has me stumped...I know (from an integral table, hah) that the integral of dx/(x^2+a^2)^1/2 is equal to ln(x+(x^2+a^2)^1/2).

    So then I'm left with

    [q/4*pi*ε*h] ln(z+(z^2+a^2)^1/2)

    However the solution is:

    V=(q/2*pi*epsilon_0 h)ln(h/2r)+sqrt{1+h^2/4r^2)

    Looks like they swapped z in terms of h, but I'm not sure where that came from?
  9. Feb 28, 2012 #8


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    Staff: Mentor

    The integration of the rings goes from -h/2 to h/2 on the z axis. Of course by symmetry you only need to integrate from 0 to h/2 and double the result. The z's in your integration will get replaced when the limits get plugged in.
  10. Feb 28, 2012 #9
    Ah, of course.

    Thank you so much for the help. This was extremely helpful and you really helped make things clear. I truly appreciate it.
  11. Feb 28, 2012 #10


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    Staff: Mentor

    My pleasure. Good luck!
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