
#1
Jan2409, 07:32 AM

P: 1

1. The problem statement, all variables and given/known data
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitude is most important along the central perpendicular axis of the disk, at a point P at distance 2.60R from the disk. Cost analysis suggests that you switch to a ring of the same outer radius R but with inner radius R/2.00. Assume that the ring will have the same surface charge density as the original disk. If you switch to the ring, by what percentage will you decrease the electric field magnitude at P? 2. Relevant equations E field equations for disk and ring 3. The attempt at a solution please tell me how to do it and the answer is ......Thanks``` 



#2
Jan2409, 10:39 AM

PF Gold
P: 363

You can define your bit of charge as [tex] dq = \rho da [/tex] where rho is the charge density per unit area. Now, for each piece of charge on the disk there is another piece diametrically opposite to it. That means that at P there will be a cancellation of components of the electric field vectors such that the resultant electric field is along the axis perpendicular to the disk. That's great because it simplifies your analysis. By now you should have figured out that you'll need to integrate over the disk to add up the charge . I'll give you the integral: [tex] E = \frac{1}{4 \pi \epsilon_o} \int \frac{dq}{r^2} [/tex] I gave you dq above, so you'll have to figure out da. You will also have to figure out how to express [tex] r^2 [/tex] in terms of the geometry. Important note: r in the equation above is the distance from the piece of charge to point P. You will have to represent the radius of the disk as you integrate with some other letter or you will mess up! Your limits of integration will depend on whether you choose to express da in terms of one variable , or two. Hint: if you choose two integrals, one of them will have limits from 0 to 2 pi. I think this will get you started. Once you understand how to calculate the electric field for the whole disk, the ring will be a piece of cake. 


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