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Homework Help: E-Field over Uniformly Charged Disc

  1. Jul 15, 2009 #1
    1. The problem statement, all variables and given/known data


    Find the Electric Field at Point P at height z over a uniformly charged disc of radius a with uniform charge Q.

    2. Relevant equations

    3. The attempt at a solution

    I'm looking at the solution online, on multiple sites, only to be stuck with the very first step.

    Why on earth is [tex]dA=2\pi*r*dr[/tex] and what exactly is dA?
  2. jcsd
  3. Jul 15, 2009 #2


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    Hi exitwound! :wink:

    (have a pi: π :wink:)

    A is area, and dA is the area of the infintesimal ring marked red in the diagram. :smile:
  4. Jul 15, 2009 #3
    I understand that concept, but I can't grasp why I take the derivative of the Area of the large disc to find the area of the small ring.
  5. Jul 15, 2009 #4
    Don't think of it like that. I'd say just think of a circular perimiter, it has circumference equal to 2piR. It's has no thickness, so no area. Now say it has an infinitesimal thickness dr. So it's area is 2piRdr. Add all these infinitesimal areas up (integrate) to get the full area.
  6. Jul 16, 2009 #5


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    Yup, nnnm4; :smile: is right … don't think of it like that!

    Yes, dA is the derivative wrt r, but that's not how you calculate dA …

    you calculate dA just by geometry … its the actual area of an actual ring :wink:

    (same when you calculate eg the volume of a solid of revolution … you divide it into very small slices, and calculate the actual volume of an actual slice)
  7. Jul 16, 2009 #6


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    I can't tell you why you would do this, but I can explain why the derivative of the area w.r.t. the radius gives you the differential area of the ring.

    POINT 1: Imagine two disks. On of them has area A, and the other one, slightly larger, has area A+ΔA. So, ΔA is the area of the ring that is formed when the first disk is subtracted from the second disk.

    POINT 2: Let the radius of the first disk be r, and the radius of the slightly larger disk be r+Δr. The area of the first disk in terms of r is A=πr^2, and the area of the second disk in terms of r is A+ΔA=π(r+Δr)^2=πr^2+2πrΔr+π(Δr)^2. So, the area of the ring is ΔA=2πrΔr+π(Δr)^2.

    POINT 3: Divide by Δr. This gives ΔA/Δr=2πr+πΔr. Now, assume that Δr<<r. Then, ΔA/Δr≈2πr, and so the area of the ring is ΔA≈2πrΔr. When we assume that Δr→0, then the approximation approaches an equality (for r>0), and the Δ's become d's: dA/dr=2πr. Now, dA is the area of a very thin ring: dA=(dA/dr)dr=2πrdr.

    In general, though, the Jacobian approach is more convenient.
  8. Jul 16, 2009 #7
    So the area of the differential ring is NOT exactly [tex]2\pi rdr[/tex]. This is what is throwing me off, I think. In my head, I read the statement [tex]dA=2\pi rdr [/tex]and I know it's not an equality. It can't be. And I think my mind hasn't been able to wrap around the fact that that statement is true as dr-->0, but not for any specific r.

    That make sense?
  9. Jul 17, 2009 #8


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    Hi exitwound! :smile:

    (what happened to that π i gave you? :wink:)
    oh that's what's worrying you!

    Well, yes you're corrrect, it's not exactly 2πrdr.

    But we're only interested in "first-order" precision …

    the error is something times (dr)2, and we're not bothered about that.

    We get a perfectly valid "first-order" equation, dA = 2πrdr, and we can bung that into the formula and integrate.

    (if you integrate (dr)2 terms, you just get zero)

    Sorry, but that's the way it works :wink: … if you're not happy with that, you'll just have to keep doing examples until you manage to convince yourself that you are happy! :smile:
  10. Jul 17, 2009 #9


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    This is the key. Think of differential quantities as something that needs to be integrated. So, as a sort of handwaving argument, imagine that dr=ε, where ε is an arbitrarily small number that can be made as small as you want compared to everything else in the problem. In other words, any product of ε with anything (except inverse powers of ε) is effectively zero.

    ∫(dr)^2 = ε∫dr

    Compared with the other integrals over r, which are not multiplied by ε, this integral is negligibly small.

    If it helps, then imagine that you will eventually want to find some numerical value. The idea is that, to however many significant figures you desire, you can always make ε small enough so that all of the terms that have an ε factor cannot add a numerical value large enough to change your numerical result in the given precision.
  11. Jul 17, 2009 #10
    Makes a little more sense now...
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