# Electric Field of a Charged Disk

by davezhan
Tags: charged, disk, electric, field
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,564 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.