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# Gauss's Law Watch

1. I'm stuck with part b. As its asking for the fractions, I feel that some symmetry principle needs to be used, so as to avoid some nasty integration to work out the explicit values of the flux's. Spent too long on this and need some help!

I do know that the flux through the end of cylinder that the charge is on, is zero, for starters.
2. (Original post by Tedward)

I'm stuck with part b. As its asking for the fractions, I feel that some symmetry principle needs to be used, so as to avoid some nasty integration to work out the explicit values of the flux's. Spent too long on this and need some help!

I do know that the flux through the end of cylinder that the charge is on, is zero, for starters.
I suppose that they want you to imagine that half of the charge is inside the cylinder and half outside. As for getting the fractions of flux, I guess that you need to consider the solid angle subtended by the various surfaces at the charge. That will give you the proportion of the total area that the flux can "shine" through.
3. (Original post by atsruser)
I suppose that they want you to imagine that half of the charge is inside the cylinder and half outside. As for getting the fractions of flux, I guess that you need to consider the solid angle subtended by the various surfaces at the charge. That will give you the proportion of the total area that the flux can "shine" through.
Thanks for that! I've had a go, but I'm struggling to find what the solid angle would be. This method also seems to be fairly complicated, and given the number of marks, I feel like there must be an easier way. If only lecturers would put up mark schemes :/
4. (Original post by Tedward)
Thanks for that! I've had a go, but I'm struggling to find what the solid angle would be. This method also seems to be fairly complicated, and given the number of marks, I feel like there must be an easier way. If only lecturers would put up mark schemes :/
Well, there's a standard formula for the solid angle subtended by a cone, which is what you'll need to find the fraction through the end of the cylinder, so you can look it up (though it's easy to derive by integration). The numbers look a bit odd though - h=30a makes for a very small solid angle. I get a fraction of
5. (Original post by atsruser)
Well, there's a standard formula for the solid angle subtended by a cone, which is what you'll need to find the fraction through the end of the cylinder, so you can look it up (though it's easy to derive by integration). The numbers look a bit odd though - h=30a makes for a very small solid angle. I get a fraction of
It is supposed to be small, but actually the answer is: for the opposite end of cylinder its 1/1800 of the flux (Q/2epsilon0). And flux through curved side is 1799/1800 of the flux. Your method makes sense so not sure why its different!
6. (Original post by Tedward)
It is supposed to be small, but actually the answer is: for the opposite end of cylinder its 1/1800 of the flux (Q/2epsilon0). And flux through curved side is 1799/1800 of the flux. Your method makes sense so not sure why its different!
OK, I screwed up before - I get using the solid angle method. You lecturer has made an approximate calculation, since the solid angle is so small we can approximate a section of a sphere by a flat surface. I'll leave you to figure out exactly what he has approximated though.
7. (Original post by Tedward)
It is supposed to be small, but actually the answer is: for the opposite end of cylinder its 1/1800 of the flux (Q/2epsilon0). And flux through curved side is 1799/1800 of the flux. Your method makes sense so not sure why its different!
I think I've got it.

If we take the electric field across the other end of the cylinder as being constant since h>>a. This gives

So the flux through the disk is
So 1/1800 of the flux goes through the end then the rest (1799/1800) goes through the curved surface
8. (Original post by atsruser)
OK, I screwed up before - I get using the solid angle method. You lecturer has made an approximate calculation, since the solid angle is so small we can approximate a section of a sphere by a flat surface. I'll leave you to figure out exactly what he has approximated though.
Lol oh well
9. (Original post by atsruser)
OK, I screwed up before - I get using the solid angle method. You lecturer has made an approximate calculation, since the solid angle is so small we can approximate a section of a sphere by a flat surface. I'll leave you to figure out exactly what he has approximated though.
(Original post by langlitz)
I think I've got it.

If we take the electric field across the other end of the cylinder as being constant since h>>a. This gives

So the flux through the disk is
So 1/1800 of the flux goes through the end then the rest (1799/1800) goes through the curved surface
Oh I see, thanks a lot! The thought of approximating never occurred, perhaps because I'm a maths student haha.
10. (Original post by Tedward)
Oh I see, thanks a lot! The thought of approximating never occurred, perhaps because I'm a maths student haha.
Yeah it's a bit of a dark art knowing when to do use an approximation, it wasn't my first thought for this problem I must admit

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