Electric Field of a Charged Disk

  1. I'm having a brain freeze and have a hard time understanding why the area of the ring is 2*pi*a*da. Can someone explain why it is the circumference times da?

    Link to derivation: www.phys.uri.edu/~gerhard/PHY204/tsl36.pdf
     
  2. jcsd
  3. berkeman

    Staff: Mentor

    How would you write the equation for the area of that ring section? What happens when you simplify what you've written?
     
  4. fluidistic

    fluidistic 3,313
    Gold Member

  5. HallsofIvy

    HallsofIvy 40,678
    Staff Emeritus
    Science Advisor

    Think of the ring as the region between two circles- one of radius a, the other of radius a+ da. The area of the inner circle is [itex]\pi a^2[/itex] and the area of the outer circle is [itex]\pi (a+ da)^2[/itex]. The area between them is [itex]\pi (a+da)^2- \pi a^2[/itex][itex]= \pi (a^2+ 2ada+ da^2)- \pi a^2[/itex][itex]= 2\pi a da+ \pi da^2[/itex]. Since da is an "infinitesmal", its square is negligible and the area is [itex]2\pi a da[/itex]. By saying that "da is an infinitesmal" I mean that this is true in the limit sense for very small da.

    Here's another way to look at it: Imagine opening that strip up to a "rectangle". It's length is the circumference of the circle, [itex]2\pi a[/itex], and it's width is da. The area of that "rectangle" is "length times width", [itex]2\pi a da[/itex]. I have put "rectangle" in quotes because, of course, you cannot "open up" a circular strip into a rectangle. This is, again, only true in the limit sense.

    If you were to take da to be any finite length, [itex]2\pi a da[/itex] would give you an approximate area, not an exact area. But you can use "da" in an integral to get the exact area.
     
  6. Thank you for your help! The above post helped to clarify things tremendously.
     
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