Electric Field of a Charged Disk

[davezhan]

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

 

[berkeman]

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

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

 

[fluidistic]

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

 

[HallsofIvy]

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 \pi a^2 and the area of the outer circle is \pi (a+ da)^2. The area between them is \pi (a+da)^2- \pi a^2= \pi (a^2+ 2ada+ da^2)- \pi a^2= 2\pi a da+ \pi da^2. Since da is an "infinitesmal", its square is negligible and the area is 2\pi a da. 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, 2\pi a, and it's width is da. The area of that "rectangle" is "length times width", 2\pi a da. 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, 2\pi a da would give you an approximate area, not an exact area. But you can use "da" in an integral to get the exact area.

 

[davezhan]

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