"Casey Hawthorne" <caseyhHAMMER_TIME@istar.ca> wrote in message
news:hbheo29mfm05hnesfre8q298v38v7a5bgm@4ax.com...
For the electric field of a uniformly charged thin ring along the axis
I get the following:
E = (1/(4 * pi * epsilon-naught))((q*z)/((R^2 + z^2)^(3/2)))
where R is the radius of the ring and the ring has charge q
You made an error somewhere. The answer should be a function of both the
inner diameter and thd outter diameter of the *annulus* (a flat disc with
smaller disk punched out of the center, e,g, a washer that would go around a
bolt).
Thanks
Pete
Starting from:
dEz = (1/(4 * pi * epsilon-naught))(q/(2*pi))(z/((R^2 +
z^2)^(3/2)))dTheta
"Pmb" <peter102560_nospam@comcast.net> wrote:
Hi folks
I need someone to double check my work. In Ohanian's EM book he has
the following question under the problem section
----------------------------------------------
A circular annulus of inner radius R1 and outer radius R2 has a
uniform charge density sigma. What is the electric field on the axis of
the
annulus at a distance z from the center?
----------------------------------------------
I worked this out and got
E = pi*sigma*z { 1/sqrt( z^2 + R2^2) - 1/sqrt( z^2 + R1^2) }
Ohanian has twice this value in his answer section. Which answer do
you get after you work this out? Thanks.
Best wishes
Pete
--
Regards,
Casey
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