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Electric Field of a Charged Disk 
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#1
Jan510, 12:39 PM

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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
Jan510, 12:57 PM

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#3
Jan510, 02:59 PM

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I don't know if it is of some help. Just in case, see post #2 : http://www.physicsforums.com/showthread.php?t=363640.



#4
Jan510, 03:38 PM

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Electric Field of a Charged Disk
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. 


#5
Jan510, 07:17 PM

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Thank you for your help! The above post helped to clarify things tremendously.



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