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Electric Fields in circular wire

140
1
I'm stumped by the following problem:

A circular ring of fine wire carries a uniformly distributed positive charge, q. Find the magnitude and direction of the electric field at the center of the ring caused by just the charge on a portion of the ring subtending an angle [the] at the center, in terms of q, [the]and radius r.

The uniform line charge [lamb] = dq/dl
l = r[the] so dl = rd[the]
dq=[lamb]rd[the]

E = kq/r^2
dE = kdq/r^2 the dE's in the y direction cancel each other out because of symmetry.
E = k[lamb]r/r^2[inte]cos[the]d[the] from -[the]/2 to [the]/2

k = 1/(4[pi][ee])

so E= [q/(2r[pi][ee]]sin([the]/2)

The book show the answer to be: [q/(4[pi]^2[ee]r^2]sin([the]/2)

Can someone please point out where I'm going wrong?

Thanks
 
1,037
1
You replaced λ with q.

Your result
E= [q/(2rΠε]sin(Θ/2)

should be
E= [λ/(2rΠε]sin(Θ/2)

Now, λ = q/(2Πr) so make that substitution & you'll get the book's answer
 
140
1
Thanks Gnome.

Boy that was simple. I guess I should read more carefully. I completely forgot that for a uniform line charge [lamb] = q/l.

Thanks again.
 

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