Helpfield of continuous distribution quest.

  • Thread starter meadow
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  • #1
meadow
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I have been staring at this problem for an hour. I am not even sure where to start.
A thin, circular disk of radius R is oriented in the xy-plane with its center at the origin. A charge Q on the disk is distributed uniformly over the surface. (a.) Find the electric field due to the disk at the point z=z0 along the z-axis.
(b.) Find the field in the limit z0 approached infinity.
(c.) Find the field in the limit that R approaches infinity.
Are the limits of parts (b) and (c) the same?

I think that the electric field due to the disk = [(Q/area)*k]*integration of the change of something over the distance squared...I am just not seeing it in simple terms...Can someone point me in the right direction?
 

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  • #2
Astronuc
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Well, you have a formula for the electric field due to a point charge, and remember that is a vector.

Now you have a field of point charges, so the field at sum point is the sum (or integral) of the vectors of each charge.

The integral involves the charge density (charge per unit area) over the area (area integral).

Are the limits of parts (b) and (c) the same?
No.

Part b asks for the field as z0 [itex]\rightarrow\,\infty[/itex], while part c asks for the field of an infinite plane, i.e. R[itex]\rightarrow\,\infty[/itex]

Let z be a position from the center of the disc, and let r be the position of a charge q from the center of the disc. That charge produces an electric field proportional to q/d2, where

d = [itex]\sqrt{z^2+r^2}[/itex]

So you should have an area integral with q/d2 in it an integrate over the area [itex]\int E_q r dr d\theta[/itex], where [itex]E_q[/itex] is the electric field contribution of one charge.

Don't forget that it is a vector, so the net electric field has a component along the z-axis, and the electric field components normal to the axis cancel.
 
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