# Electric Field of a Charged Disk

by davezhan
Tags: charged, disk, electric, field
 P: 3 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
Mentor
P: 39,648
 Quote by 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
How would you write the equation for the area of that ring section? What happens when you simplify what you've written?
 PF Gold P: 3,173 I don't know if it is of some help. Just in case, see post #2 : http://www.physicsforums.com/showthread.php?t=363640.
Math
Emeritus
Thanks
PF Gold
P: 38,890

## 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 $\pi a^2$ 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.
 P: 3 Thank you for your help! The above post helped to clarify things tremendously.

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