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- Thread starter david90
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chroot

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What is the total charge? Obviously, it's just the circumference of the circle times its charge density:

Q = 2 pi R [rho]

= 2 pi (1) (1)

= 2 pi coulombs

However, you could also obtain the same result by integrating. What you're going to do conceptually is break down the circle into a lot of little tiny pieces, each with it's own little tiny charge, and then sum up the charge over all the little tiny pieces. In the limit as the pieces get very small, you're doing an integration.

So, let's investigate a small piece of the circle. If an arc length is called s, a little tiny piece of an arc length is called ds. If charge is called q, then a little tiny piece of charge is called dq. How much charge does a tiny piece of the circle have on it?

dq = [rho] * ds

It has a charge equal to the linear charge density [rho] times its length, ds.

Now we can integrate all these little tiny pieces of charge over the whole arc length of the circle (2 pi R), like this:

Integral from 0 to (2 pi R) of ([rho] ds)

which is just 2 pi R [rho], exactly as we found before.

Does this make sense? Let me know if anything is confusing.

- Warren

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turin

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chroot

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That's true, most integrals in physics are over some spatial dimension(s).Originally posted by turin

You need to get dq in terms of dx

- Warren

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which is just 2 pi R P, exactly as we found before.

I don't see how u go from 0 to 2pi R |p ds to 2pi R P.

Did u integra with respect to ds? Dont u have to change it to

something else?

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which is just 2 pi R P, exactly as we found before.

I don't see how u go from 0 to 2pi R |p ds to 2pi R P.

Did u integra with respect to ds? Dont u have to change it to

something else?

Btw, U did turn on a light bulb inside my head. it is just a little

dim.

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chroot

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Yes, I integrated with respect to ds. What's the value of this integral:Originally posted by david90

I don't see how u go from 0 to 2pi R |p ds to 2pi R P.

Did u integra with respect to ds? Dont u have to change it to

something else?

Integral from 0 to x of (ds)?

It's just x, of course. The same thing applied here. [rho] is a constant and can be pulled out of the integral.

- Warren

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