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Homework Help: Electric Field due to a charged hallow cylinder/solid cylinder on a point

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a uniformy charged thin-walled right circular cylindrical shell having total charge Q, radius R, and height h. Determine the electric field at a point a distance d from the right side of the cylinder (treat the cylinder as a collection of ring charges). Consider now a solid cylinder with the same dimenstions carrying the same charge,uniformly distrubted through its volume (treat the cylinder as a collection of disk charges)


    2. Relevant equations
    Ring Electric Field: [tex]\int \frac{kx dq}{(x^{2} + a^{2})^{\frac{3}{2}}} [/tex]

    Disk Electric Field: [tex]\int \frac{kx\pi\sigma 2rdr}{(x^{2} + a^{2})^{\frac{3}{2}}}[/tex]

    The disk equation is derived from the ring equation by treating a disk as a series of rings of infinitesimal radius and integrating using the substition [tex]dq = \pi\sigma 2rdr[/tex]

    3. The attempt at a solution
    For the first portion of the problem I started with the Ring Equation and used the equation [tex]dq = \lambda dx[/tex]. This gave me
    [tex]k\lambda\int \frac{xdx}{(x^{2} + R^{2})^{\frac{3}{2}}}[/tex]
    Using the substitution [tex] u = (x^{2} + R^{2}), du = 2xdx [/tex] I have
    [tex]\frac{k\lambda}{2}\int \frac{du}{u^{\frac{3}{2}}} = \frac{-k\lambda}{\sqrt{u}}|^{d+h}_{d} = k\lambda(\frac{1}{\sqrt{d^{2} + R^{2}}} - \frac{1}{\sqrt{(d+h)^{2} + R^{2}}})[/tex]

    I went online to compare my results with others and found someone had asked the https://www.physicsforums.com/showthread.php?t=188011" as myself. My answer seems to follow their line of logic. However I can't get an appropriate answer for the 2nd part of the problem. It seems to me like I'd just take the same approach and just plug [tex] dq = \rho dV =\rho\pi r^{2}dx [/tex] into the Disk equation. However they replaced [tex] dq [/tex] already, and if I go back and plug my [tex] dq [/tex] into the ring equation (like the book did with it's [tex]dq[/tex]) then I just end up with the same equation as above. Specifically I get

    [tex]\rho k\pi R^{2}(\frac{1}{\sqrt{d^{2} + R^{2}}} - \frac{1}{\sqrt{(d+h)^{2} + R^{2}}})[/tex].

    But since [tex] \rho = \frac{Q}{V} = \frac{Q}{\pi R^{2} h} [/tex] and [tex] \lambda = \frac{Q}{h} [/tex] both equations reduce down to

    [tex]\frac{kQ}{h}(\frac{1}{\sqrt{d^{2} + R^{2}}} - \frac{1}{\sqrt{(d+h)^{2} + R^{2}}})[/tex].

    Any help is greatly appreciated
     

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    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Jan 17, 2010 #2
    Anybody?
     
  4. Jan 17, 2010 #3

    vela

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    Your expression for the disk is wrong. You seem to understand the approach to solve the problem conceptually, but you don't seem to get the math quite yet. I'll show how you'd go about getting the electric field for the disk. You want to follow a similar approach to solve the cylinder problem.

    In the integral for the electric field due to a charged ring of radius a,

    [tex]
    \int \frac{kx dq}{(x^{2} + a^{2})^{\frac{3}{2}}}
    [/tex]

    [itex]dq[/itex] represents the charge of an infinitesimal piece of the ring. The rest of the integrand is constant, so you get

    [tex]
    \int \frac{kx dq}{(x^{2} + a^{2})^{\frac{3}{2}}} = \frac{kx}{(x^{2} + a^{2})^{\frac{3}{2}}} \int dq = \frac{kxQ}{(x^{2} + a^{2})^{\frac{3}{2}}}
    [/tex]

    where Q is the total charge on the ring. So far so good.

    Now when you calculate the electric field of a disk with uniform charge density [itex]\sigma[/itex], you can think of the disk as a collection of rings. A ring of radius r and thickness dr has a charge [itex]dq=\sigma 2\pi r dr[/itex], so its contribution to the electric field will be

    [tex]
    dE = \frac{kxdq}{(x^{2} + r^{2})^{\frac{3}{2}}} = \frac{kx(2\pi \sigma r dr)}{(x^{2} + r^{2})^{\frac{3}{2}}}
    [/tex]

    Note that the radius of the ring is no longer equal to the constant a. To get the total electric field of the disk, you'll now have to integrate. Once you have that result, then you can use it to calculate the field due to the solid cylinder, modeling the cylinder as a collection of disks.
     
  5. Jan 17, 2010 #4
    I understand that logic (at least I think I do), actually my textbook goes through the problem of representing a disk as a series of rings, but I couldn't gather how to apply that approach to a cylinder (thus my confusion). Specifically my book says:

    [tex]\int \frac{kx\pi\sigma 2rdr}{(x^{2} + r^{2})^{\frac{3}{2}}} = 2\pi k\sigma(1 - \frac{x}{(x^{2} + r^{2})^{1/2}})[/tex].

    However in all the previous problems they have gone back to a more basic equation involving [tex] dq [/tex] and substituted either [tex] dq = \lambda dx [/tex] or [tex] dq = \sigma dA[/tex] or [tex] dq = \rho dV [/tex] into the integral. Since the disk equation does not contain a [tex] dq [/tex] I can't see how to apply the previous examples to this problem. I assume its not as simple as taking the disk equation and sticking a [tex] dx [/tex] onto the end of it
     
    Last edited: Jan 17, 2010
  6. Jan 17, 2010 #5

    vela

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    I wanted to make a point about your calculation for the hollow cylinder. I think your answer is correct, but that was an accident. Your method happened to work because the expression for the electric field of the ring happens to look like the original integrand with dq replaced by Q. The correct method is to start with the integrated expression, replace Q with dq for the cylinder, and then integrate.
     
  7. Jan 17, 2010 #6
    Wow that is a fluke, Thanks a lot I really appreciate it!
     
  8. Jan 17, 2010 #7

    vela

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    That can't be right because you have an [itex]a[/itex] on the lefthand side, and it doesn't appear on the righthand side.

    Do you understand how the book came up with original integrands? If you get that, the problems are relatively straightforward.
     
  9. Jan 17, 2010 #8
    Oh thats my bad that a should be a r, in my haste when making the forum post I just copied the Ring intergral and changed it into the one for the disk. I apperently missed that :blushing:

    Yeah now I see how it all fits together, I just happened to make the fluke and mess up the rest because of it.
     
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