Calculating E in a hemispherical shell using Coulomb's law

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The discussion centers on calculating the electric field (E) in a hemispherical shell using Coulomb's law. The teacher's method involves slicing the hemisphere into small rings, with the electric field along the axis being directed entirely in the z-direction due to symmetry. A participant provides a scanned diagram and explanation of the geometry involved, which relates the electric field from a ring of charge to the overall field in the hemisphere. The shared resources aim to clarify the calculation process for others struggling with the concept. Understanding the geometry and symmetry is crucial for solving the problem effectively.
Lisa...
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My teacher worked out the following problem as shown below:

http://img116.imageshack.us/img116/7656/naamloos27mf.gif

The only thing is I don't really understand what he's doing, so could anybody please explain to me a bit more which steps he takes? I would be very greatful! (or maybe there's an easier way to solve it...)
 
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Basically what he's doing is slicing the hemisphere into a whole bunch of tiny rings. Presumably equation 23-10 describes the electric field generated by a ring along the axis which runs through its center. Due to symmetry in x and y (there's just as much charge to one side of the ring as the other) the electric field along this axis must be entirely in z.

The rest of it is just some geometry on the surface of the sphere, unfortunately I don't have time to draw a picture and scan it just now. If nobody else has replied by the time I get back from class I'll see if I can whip something up for you.
 
Hi Lisa,

I wrote up the problem and scanned it, hopefully this will help. I derived the E field from a ring of charge up near the top since I didn't have a handy reference. I then drew out the geometry on the surface of the sphere and used that to relate back to the field from the ring of charge. Hopefully this will help you see what's going on.

The file is http://www.warpenguin.com/~dice/hemisphere.jpg , it's huge though. I think you should be able to print it out and it should look okay, or you can just zoom out and read it on your monitor.

Let me know if you have any questions :smile:
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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