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vorcil
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Homework Statement
A thin, circular ring of inner radius a and outer radius b carries a uniform surface charge density [tex] \sigma [/tex]
(i) Find an expression for E at a point on an axis perpendicular to the plane of the disk, the axis passes through the centre of the disk.
(ii) Keeping the surface charge density the same as that in part (i) find the electric field, E, as a->o and b->infinity
Homework Equations
The usual ones, e.g [tex] \frac{1}{4\pi \epsilon_o} \frac{q}{r^2}\hat{r} [/tex]
The Attempt at a Solution
I can't really show you what I've drawn for a picture,
but imagine a circular disk, with the normal being the z axis, with a hole in the middle,( like a dvd disk)
where the distance from the center of the disk to the inner part is radius a, and to the edge of the disk, radius b
solving for the Electric field:
where sigma is the charge surface density of the disk
and R is the radius
and varsigma [tex] \varsigma [/tex] [tex] \sqrt{R^2 + z^2 } [/tex]
and [tex] cos \theta = \frac{R}{\varsigma} [/tex][tex] dE = \frac{1}{4\pi \epsilon_o} \sigma \frac{1}{\varsigma^2} cos \theta \hat{z} [/tex]
I think I also have to multiply the equation by [tex]2\pi R[/tex] but I'm not too sure why, it's just intuition,
making it
[tex] dE = \frac{2\pi R}{4\pi \epsilon_o} \sigma \frac{1}{\varsigma^2} cos \theta \hat{z} [/tex]
substituting in the varsigma and the cos theta,
[tex] dE = \frac{2\pi R }{4\pi \epsilon_o} \sigma \frac{1}{R^2+z^2} \frac{R}{\sqrt{R^2+z^2}}\hat{z} [/tex]
integrating(and taking out the constant)
[tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \int \frac{1}{R^2+z^2} \frac{z}{\sqrt{R^2+z^2}}\hat{z}[/tex]
[tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \int \frac{1}{R^2+z^2} \frac{R}{\sqrt{R^2+z^2}}\hat{z}[/tex]
[tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \int \frac{1}{(R^2+z^2)^{\frac{3}{2}}}[/tex]
[tex] \bf{E} = \frac{2\pi R }{4\pi \epsilon_o} \sigma \frac{R}{z^2\sqrt{R^2+z^2}} |_a^b[/tex]
now I'm pretty sure this is the general result for a disk, not a disk with a hole,
but I've been given the inner and outer radii, so I think I can just take the two disks and subtract the smaller disk from the larger disk to get the electric field of the ring (because electric fields and potential can be super imposed)
giving me the final expression for the electric field strength E, at a point on the Z axis (in the normal direction)
[tex] \bf{E} = \frac{R \sigma}{2\epsilon_o} \frac{R}{z^2\sqrt{R^2+z^2}}|_a^b[/tex]
=
[tex] [tex] \bf{E} = \frac{\sigma}{2\epsilon_o} (b-a) \frac{b-a}{z^2\sqrt{(b-a)^2+z^2}}[/tex]
can someone please tell me if that's an accurate expression for the electric field of a ring?
cheers
for part(b)
letting a->0 and b->infinity,
in the limits, it becomes a singular disk without a hole and an infinite radius
[tex] \bf{E} = \frac{\sigma}{2\epsilon_o} \frac{1}{z^2}[/tex] since all the r's cross out, this is the equation for a disk of infinite radius assuming my part(A) was correct
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