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Electric field of a uniformly charged disk

by Woopydalan
Tags: charged, disk, electric, field, uniformly
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Woopydalan
#1
Feb8-13, 03:52 PM
P: 746
Hello,

I am looking at an example of finding the charge of a uniform disk with a continuous charge on the surface.

They go about the problem by finding the infinitesimal charge of concentric rings

dq = σdA = σ(2πr dr)

The part I don't understand is that they use the area as 2πr dr? The area of a ring would be ∏(R2^2 - R1^2), right?
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jtbell
#2
Feb8-13, 04:05 PM
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Quote Quote by Woopydalan View Post
The area of a ring would be ∏(R2^2 - R1^2), right?
Correct, that's the exact area of the ring. However, if dr = R2 - R1 is small, the other formula is a very good approximation, which gets better and better as dr becomes smaller and smaller. To see this, let R2 = R1 + dr in the equation above, cancel out whatever you can, and then drop any terms with (dr)^2. Those terms become negligible compared to terms with just dr, when dr is very small.
Woopydalan
#3
Feb8-13, 04:14 PM
P: 746
Ok I just calculated it and now I see. The book didn't even bother to explain that approximation, which left me confused. Thank you for clearing that up so quickly =)


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