Conductors have always spun me in circles intuitively. So, |Egap|=|σ1|/εo=|σ2|/εo, if I'm correct. This could be concluded using the boundary condition that ΔE=σ/εo when crossing a surface, or by using Gauss' Law.Delta2 said:Given that s<<R we can make the approximation that all σiσi will be constant throughout their respective domains. Also that the two spherical caps each carry charge +Q/2 since the whole system is very similar to an uncut sphere (because s<<R). This leaves that the flat face of the upper hemisphere (to which we put total charge +Q) has charge +Q/2 and the flat face of the lower hemisphere has charge -Q/2. Again because s<<R the electric field due to the charges on the flat surfaces will be zero in the space that is not in-between the flat surfaces. The electric field in-between the flat surfaces will be that of a parallel plate capacitor with charge Q/2 and -Q/2 in each plate.
I think with the guide of @TSny the problem has been solved, you just haven't realized it :D.
Yes and ##\sigma_1=?##Tyler DeFrancesco said:Conductors have always spun me in circles intuitively. So, |Egap|=|σ1|/εo=|σ2|/εo, if I'm correct.
Delta2 said:Yes and ##\sigma_1=?##
Tyler DeFrancesco said:Also, σ1=Qeff/A=(1/2)Q/(πR2).
Awesome. Thank a lot for the help, the both of you. Electromagnetism has always been more painful for me than any other field of physics so far in my education. Even quantum 1 (with Griffiths) isn't as perturbing.Delta2 said:Yes correct.
You are welcome ! :DTyler DeFrancesco said:Awesome. Thank a lot for the help, the both of you. Electromagnetism has always been more painful for me than any other field of physics so far in my education. Even quantum 1 (with Griffiths) isn't as perturbing.
No you are not doing the proper approximation with this reasoning.The final question is to find the potential difference between the hemispheres. V should be constant everywhere but the gap. Since s<<R, can I integrate the entire thing as a spherical shell? If so, I reckon I would get a difference of zero, since V would be constant and equal everywhere on the surface if we neglect the gap entirely.
Right, that's about where I am. Would integrating ∫ρ dv across the gap (as a 3D disc using cylindrical coordinates) work? I would find volume charge of the space by using Gauss Law in differential form.Delta2 said:You are welcome ! :D
No you are not doing the proper approximation with this reasoning.
Each hemisphere is a conductor in static equilibrium so it has a constant V through out its surface. The potential on the upper spherical cap is the same as the potential of the upper flat surface, and the potential of the lower spherical cap is the same as that of the lower flat surface (though they carry opposite charges). But each hemisphere has different V. The potential difference between the hemispheres is the potential difference of the parallel plate capacitor formed in the gap.
I'm not exactly sure how that might be done based on what you've said.Delta2 said:yes well sorry its not exactly a parallel plate capacitor its a capacitor of two parallel circular disks, but the formula for the electric field in between remains as we said at post #36.
You just have to integrate the electric field over the gap s, to find the potential difference (instead of doing it the hard way and integrate ##\frac{\rho(r')}{|r-r'|}dV'## and I don't know if this is possible because here we have surface charge densities not volume charge densities).
you have the electric field ##E=\frac{\sigma_1}{\epsilon_0}##. Isn't it straightforward to integrate and compute ##V=\int_0^s E\cdot dl##Tyler DeFrancesco said:I'm not exactly sure how that might be done based on what you've said.
Yes, it came out to be V=Es, as I mentioned in post #42, should I choose dl to run along s. This is clear even beforehand, as E is not dependent on s, right?Delta2 said:you have the electric field ##E=\frac{\sigma_1}{\epsilon_0}##. Isn't it straightforward to integrate and compute ##V=\int_0^s E\cdot dl##
yes I guess you were editing post #42 cause it initially was much shorter.Tyler DeFrancesco said:Yes, it came out to be V=Es, as I mentioned in post #42 should I choose dl to run along s. This is clear even beforehand, as E is not dependent on s, right?
Substituting in E=q/2πεoR2 gives the full answer, obviously (if that's the extra information you were asking for.
Ah. Thank you for the clarification, and again for the overall help on the problem. At least in classical mechanics, there are still limiting cases that I could check for consistency. In electromagnetism, much of the time I feel as if I cannot even make sense of the limiting cases themselves as they are not always intuitive in the same sense. In my current bracket of quantum, everything is obviously so mathematical and statistical that the physical analogue is not needed for many cases thus far, and the class is a little less second-guessy all the time.Delta2 said:yes I guess you were editing post #42 cause it initially was much shorter.
So yes I believe the potential difference is ##V=\frac{Qs}{2\pi\epsilon_0R^2}##
Yes well , this problem with the limit of s<<R makes things a bit unclear on what are the proper approximations to make. Also things appeared to become a bit contradictory in the course of events as we say that the system is similar to an uncut sphere (which has the same potential in the two hemispheres) but then we say there is a potential difference between the hemispheres. But if s is small compared to R then the potential difference which we found to be proportional to ##\frac{s}{R^2}## will also be small, and the system is indeed similar to an uncut sphere.Tyler DeFrancesco said:Ah. Thank you for the clarification, and again for the overall help on the problem. At least in classical mechanics, there are still limiting cases that I could check for consistency. In electromagnetism, much of the time I feel as if I cannot even make sense of the limiting cases themselves as they are not always intuitive in the same sense. In my current bracket of quantum, everything is obviously so mathematical and statistical that the physical analogue is not needed for many cases thus far, and the class is a little less second-guessy all the time.
The final result's consistency with the seemingly-contradictory assumptions we've made is a good point, thank you for pointing that out.Delta2 said:Yes well , this problem with the limit of s<<R makes things a bit unclear on what are the proper approximations to make. Also things appeared to become a bit contradictory in the course of events as we say that the system is similar to an uncut sphere (which has the same potential in the two hemispheres) but then we say there is a potential difference between the hemispheres. But if s is small compared to R then the potential difference which we found to be proportional to ##\frac{s}{R^2}## will also be small, and the system is indeed similar to an uncut sphere.