Electrodynamics: Conducting sphere of radius R cut in half

[TSny]

I’m not understanding how that reveals information about the planar faces of the hemispheres. They don’t exist in a non-cut sphere analogue, so what happens to the charge on that?

You'll be able to deduce the charge on the planar faces after you see how much charge is on each hemispherical cap. But you do already know that the net charge of both planar surfaces is zero since they carry equal but opposite charge.

What's the total charge of the entire system?

What's the total charge on the two hemispherical caps?

{supper time}

 

[sdefresco]

You'll be able to deduce the charge on the planar faces after you see how much charge is on each hemispherical cap. But you do already know that the net charge of both planar surfaces is zero since they carry equal but opposite charge.

What's the total charge of the entire system?

What's the total charge on the two hemispherical caps?

{supper time}

Total charge of the system is +Q.
I don’t know what the total charge on the caps is. +Q or some fraction of Q dictated by geometry, I suppose.
They need to net to Q, so 1/2 Q each sounds right. I feel the caps’ charge need to be opposite in order for E=0 to be maintained within each of the hemispheres, however.

 

[TSny]

Total charge of the system is +Q.
I don’t know what the total charge on the caps is. +Q or some fraction of Q dictated by geometry, I suppose.
They need to net to Q, so 1/2 Q each sounds right.

Right. Each cap has a net charge of Q/2, where Q is the amount of charge that was initially placed on the upper hemispere.

I feel the caps’ charge need to be opposite in order for E=0 to be maintained within each of the hemispheres, however.

Remember the basketball. You had to place a uniform charge density over the entire surface in order to get E = 0 everywhere inside. So, the charge density for each hemispherical cap must be the same (including sign).

 

[sdefresco]

Right. Each cap has a net charge of Q/2, where Q is the amount of charge that was initially placed on the upper hemispere.

Remember the basketball. You had to place a uniform charge density over the entire surface in order to get E = 0 everywhere inside. So, the charge density for each hemispherical cap must be the same (including sign).

It seems a bit counter intuitive that this could be true for electrically isolated objects. This would mean that σ₁ and σ₃ have the same sign, but that's impossible since E=0 would not be true in this case for the hemisphere. I feel like calling the objects "electrically isolated" (as specified in the prompt), and then saying s<<R, doesn't actually give me anything to go off of, but rather only further confuses me due to these questions. I could say I know how everything will re-arrange themselves by concluding E=0 in a conductor, but that's not exactly true because clearly it's been days and I still don't have a solid answer. If I'm at a constant loss to really justify that my physical system follows my train of logic, I won't gain anything out of this.

Frankly, I have to turn this problem set in within a couple of hours, and still have nothing worth writing on the page.

 

[Delta2]

It seems a bit counter intuitive that this could be true for electrically isolated objects. This would mean that σ₁ and σ₃ have the same sign, but that's impossible since E=0 would not be true in this case for the hemisphere.

Given that s<<R we can make the approximation that all $\sigma_i$ 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, because the two flat surface each with equal and opposite charge are like a parallel plate capacitor. 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.

Frankly, I have to turn this problem set in within a couple of hours, and still have nothing worth writing on the page.

I think with the guide of @TSny the problem has been solved, you just haven't realized it :D.

 

[sdefresco]

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.

Conductors have always spun me in circles intuitively. So, |E_gap|=|σ₁|/ε_o=|σ₂|/ε_o, if I'm correct. This could be concluded using the boundary condition that ΔE=σ/ε_o when crossing a surface, or by using Gauss' Law.

Also, σ₁=Q_eff/A=(1/2)Q/(πR²).

 

[Delta2]

Conductors have always spun me in circles intuitively. So, |E_gap|=|σ₁|/ε_o=|σ₂|/ε_o, if I'm correct.

Yes and $\sigma_1=?$

 

[sdefresco]

Yes and $\sigma_1=?$

Also, σ₁=Q_eff/A=(1/2)Q/(πR²).

 

[Delta2]

Yes correct. This problem is one of those problems that you got to understand which approximations to make and then the solution reveals itself.

If you can't understand what approximations to make, then the only thing that saves you is to use a numerical solver for PDEs to solve Poisson's equation with boundary conditions..

 

[sdefresco]

Yes correct.

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.

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.

 

[Delta2]

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.

You are welcome ! :D

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.

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.

 

[sdefresco]

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.

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.

If that is no good, I assume the line integral would be the next place to go (E dot dl). In that case, would my path of integration simply be a line from the upper to the lower hemisphere? Something seems odd about me taking a line integral over a 3D field (although it is uniformly in one direction).

The line integral is telling me ΔV=Es (since E has no s dependence and is uniform and parallel to s at all points along path). This result makes sense, I suppose - the suggested proportionality, at least.

 

[Delta2]

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).

 

[sdefresco]

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).

I'm not exactly sure how that might be done based on what you've said.

 

[Delta2]

I'm not exactly sure how that might be done based on what you've said.

you have the electric field $E=\frac{\sigma_1}{\epsilon_0}$. Isn't it straightforward to integrate and compute $V=\int_0^s E\cdot dl$

 

[sdefresco]

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, 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/2πε_oR² gives the full answer, obviously (if that's the extra information you were asking for.

 

[Delta2]

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/2πε_oR² gives the full answer, obviously (if that's the extra information you were asking for.

yes I guess you were editing post #42 cause it initially was much shorter.
So yes I believe the potential difference is $V=\frac{Qs}{2\pi\epsilon_0R^2}$

 

[sdefresco]

yes I guess you were editing post #42 cause it initially was much shorter.
So yes I believe the potential difference is $V=\frac{Qs}{2\pi\epsilon_0R^2}$

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]

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.

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.

 

[sdefresco]

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.

The final result's consistency with the seemingly-contradictory assumptions we've made is a good point, thank you for pointing that out.