Question about the E-field along center axis of charged ring

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SUMMARY

The discussion focuses on understanding the derivative notation "dE" and "dQ" in the context of calculating the electric field along the center axis of a charged ring. Participants clarify that "dq" represents the charge on an infinitesimally small segment of the ring, which is integrated to find the total electric field. The standard procedure involves dividing the charged body into small slices, allowing for constant measurements within those slices. The total electric field is computed using the formula ∫ kxdq/(x²+a²)^(3/2).

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Loopas
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http://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html

I'm having trouble understanding the derivative notation. I know that "dE" and dQ" have to do with the derivative, but what exactly does that mean in the context of this problem? For some reason I find it difficult to understand why this derivative notation is used in so many physics questions. How do I know what variables have derivative notation?
 
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Hi Loopas! :smile:
Loopas said:
I'm having trouble understanding the derivative notation. I know that "dE" and dQ" have to do with the derivative, but what exactly does that mean in the context of this problem? … How do I know what variables have derivative notation?

It isn't a derivative … in fact, you're going to integrate it.

dq is the charge on a tiny piece of the ring.

This is a standard procedure: you split the body into slices that are so small that you can regard all measurements on them as constant.

(for example, if you know the density of a body as a function of position, you say consider a small element of volume dxdydz, it can be assumed to have a constant density ρ, so the mass is ρdxdydz, and the total mass is ∫ ρdxdydz)

In this case, the charge is dq, the total charge is ∫ dq, the Ex for that charge is kxdq/(x2+a2)3/2, and so the total Ex is ∫ kxdq/(x2+a2)3/2 :wink:


(btw, in this case, all measurements on the small charge really are constant, so ther's actaully no assumption here … but usually there is)
 

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