Register to reply

Electric Field of a Charged Disk

by davezhan
Tags: charged, disk, electric, field
Share this thread:
davezhan
#1
Jan5-10, 12:39 PM
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
Phys.Org News Partner Physics news on Phys.org
Symphony of nanoplasmonic and optical resonators produces laser-like light emission
Do we live in a 2-D hologram? New Fermilab experiment will test the nature of the universe
Duality principle is 'safe and sound': Researchers clear up apparent violation of wave-particle duality
berkeman
#2
Jan5-10, 12:57 PM
Mentor
berkeman's Avatar
P: 41,012
Quote Quote by davezhan View Post
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
#3
Jan5-10, 02:59 PM
PF Gold
fluidistic's Avatar
P: 3,201
I don't know if it is of some help. Just in case, see post #2 : http://www.physicsforums.com/showthread.php?t=363640.

HallsofIvy
#4
Jan5-10, 03:38 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,524
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.
davezhan
#5
Jan5-10, 07:17 PM
P: 3
Thank you for your help! The above post helped to clarify things tremendously.


Register to reply

Related Discussions
Uniformly-charged disk electric field at ANY point (i.e., off-axis)? Advanced Physics Homework 2
The Electric Field Due to a Charged Disk Introductory Physics Homework 1
Electric Field Due to a Charged Disk Introductory Physics Homework 1
Electric Field due to a Charged Disk. Introductory Physics Homework 2
Approximation of electric field of uniform charged disk Introductory Physics Homework 3