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Calculating area of an annulus

  1. May 30, 2013 #1
    My physics book when providing proof of the electric potential of a disk tells me that the area of a ring with radius [itex]R'[/itex] and width [itex]dR'[/itex] is [itex]2 \pi R'dR'[/itex]
    ringofcharge.JPG
    The problem is, I have no idea how he arrives at this conclusion. Here is my attempt:
    [itex]A = \pi (R'+dR')^2 - PiR'^2\\ A = \pi(R'^2 + dR'^2 + 2R'dR') - \pi R'^2 \\ A = \pi dR'^2 + 2 \pi R' dR'[/itex]
    Which is, of course, different from what my book says:
    [itex]A = \pi dR'^2 + 2 \pi R' dR' \neq 2 \pi R'dR' [/itex]
     
  2. jcsd
  3. May 30, 2013 #2
    Notice that [itex]dR'^2[/itex] is the square of an infinitesimal quantity, which we take to be zero. So your derivation works out.
     
  4. May 30, 2013 #3

    Office_Shredder

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    When doing infinitesimal proofs like this any squared terms are negligible - for example your [itex] dR'^2[/itex] should be thrown out.

    An alternate way to see the result is if you cut the ring and lay it out flat it's a rectangle (almost) with side lengths easily calculated.
     
  5. May 30, 2013 #4

    LCKurtz

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    I will drop the primes so as not to confuse with the derivative. What your text is using is the differential approximation to the annulus area. You have a circle of area ##A=\pi R^2##. The approximate change in area when ##R## is changed by an amount ##dR## is the differential ##A'dR = 2\pi R dR## which approximates the exact change, which you have calculated. For small ##dR## you can ignore the ##dR^2## term.

    [Edit] Wow! Two other answers just while I was typing. That's life in PF.
     
  6. May 30, 2013 #5
    Ohh, I see ! Thanks to you three, that perfectly cleared my doubt.
     
  7. May 31, 2013 #6

    HallsofIvy

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    A disk of radius R has area [itex]\pi R^2[/itex]. A disk of radius R+ dR has area [itex]\pi(R+ dR)^2= \pi R^2+ 2\pi RdR+ \pi(dR)^2[/itex]. The area of the annulus between them is [itex]2\pi RdR+\pi (dR)^2[/itex]. If dR is much smaller than R, that is approximatelyu [itex]2\pi RdR[/itex].
     
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