At 12:30, you say that E=0 everywhere inside the conductor follows from the charge inside the sphere being zero. Isn't the implication only true in the opposite direction. Surely the integral of E over the surface can evaluate to zero q and the integrand E still possibly be non-zero in places inside? Now, of course if E is zero everywhere inside, the enclosed q will be zero (the implication in the other direction). My question: The usual proof that the E field is zero for an electrostatic arrangement of charge on a solid conducting sphere in books is by contradiction: It basically says that in an electrostatic arrangement, the E field has to be zero everywhere inside, since if it were non-zero anywhere, we could place a charge at that point inside and we would no longer have an electrostatic arrangement. But isn't placing a charge inside changing the original electrostatic arrangement that we are supposedly contradicting. The original electrostatic arrangement (without the additional charge placed inside) may well be still be consistent with an E field inside at a particular point. At least we haven't contradicted the original electrostatic arrangement, have we? You might be able to shed light on this. Thanks.
Hi, I'm a little confused about what you were saying about "the opposite direction" and "contradiction". Could you explain that again please? As for the electric field, we say it's zero inside the sphere - not because there is no electric field, but because all the field lines are pointing in all sorts of directions so that they end up canceling out to zero. So when you say that the electric field could be nonzero somewhere inside the sphere, I don't think that's right (but I could be wrong). Then, if you were to add a charge (assuming positive) inside the sphere, my guess is that the charge would immediately go to the outer edge of the sphere so that the charge enclosed is still zero.
@@danthetutor2624 Hi thanks for your reply. All I mean by implication in the "opposite direction" is: 1. If E=0 everywhere inside, then the integral of E over the surface is zero, then q_enclosed is zero (one direction of implication). 2. But, if q_enclosed is zero, this doesn't mean E everywhere inside has to be zero (implication in the opposite direction is false; the converse is not true essentially). This is just from the mathematics: if the integrand E is zero, the integral (evaluated as q_enclosed) is zero but if the integral is zero, the integrand E doesn't have to be zero everywhere. Regarding your second paragraph, I'm also talking about E being the net electric field at any point inside: I think we are on the same page about that. As you say, it is proposed that the electric field contributions from all the charges cancel at every point inside i.e. that E=0 everywhere inside. I'm not saying E is zero or E is not zero: I'm just saying we can't find whether E is or is not zero just from the fact that q_enclosed=0, as seems to have been done here. Anyway, whatever about that, my main question has to do with how this E=0 is arrived at. Books arrive at it by "proof by contradiction" (that's what I mean by contradiction above). So it typically goes like: 1. Take an electrostatic arrangement of charge on a conducting solid sphere. 2. The net E field everywhere inside must be zero. 3. For If it wasn't zero at a point inside, we could place a charge at that point and obviously the charge would start moving. 4. This moving charge contradicts our electrostatic arrangement assumption above. 5. So we can't have a non-zero net E at any point inside. My issue with this is step 3 seems to be a (slightly) different electrostatic arrangement of charge to step 1. We've changed the situation slightly by placing an extra charge inside. So this usual argument in textbooks seems flawed. How I've tried to resolve it is by assuming that E fields (and E field lines) are continuous in free space. So if we had a non-zero E field at any point inside, there will an (attenuated) extra non-zero E field where the actually charges are themselves, moving them and changing the electrostatic arrangement. Which is a contradiction. So the net E field inside everywhere is zero. I think I can accept that E fields are continuous in free space (by Gauss's law) so my last paragraph makes sense to me. The steps above (how it's presented in textbooks) doesn't quite make sense. Can you comment on any of that. Thanks
Ok, I see what you're saying. I hate trying to read textbook explanations. I also don't think I can add anything to what you've already said because that's about my understanding of it. If you're in school or university right now, I would probably ask one of your teachers.
@@danthetutor2624 Ah that's fine. That's for getting back. I'm always just trying to refine my understanding and check if people like yourself might know something that I'm just not getting. It can be frustrating, especially if it's probably something insignificant like just misinterpreting a poorly worded textbook explanation as you say. But cheers anyway. Appreciate your help.
You can just treat that as r>a. A gaussian drawn on the surface of the sphere would include all the charge and be no different from any other gaussian drawn further out ;)
Thank you dan, I hope your high tea went well, also i might actually pass physics now. This really wasnt as complicated as I thought
That tea was nearly 2 years ago, but yes it went well. Glad to hear you might pass physics!
thank you for saving me
No prob. Good luck on your exam!
It is helpful to me thank you 🙏☺️
Thanks for watching! Happy to help!
@@danthetutor2624 in conducting sphere if r= a ?
At 12:30, you say that E=0 everywhere inside the conductor follows from the charge inside the sphere being zero.
Isn't the implication only true in the opposite direction.
Surely the integral of E over the surface can evaluate to zero q and the integrand E still possibly be non-zero in places inside?
Now, of course if E is zero everywhere inside, the enclosed q will be zero (the implication in the other direction).
My question:
The usual proof that the E field is zero for an electrostatic arrangement of charge on a solid conducting sphere in books is by contradiction:
It basically says that in an electrostatic arrangement, the E field has to be zero everywhere inside, since if it were non-zero anywhere, we could place a charge at that point inside and we would no longer have an electrostatic arrangement.
But isn't placing a charge inside changing the original electrostatic arrangement that we are supposedly contradicting. The original electrostatic arrangement (without the additional charge placed inside) may well be still be consistent with an E field inside at a particular point. At least we haven't contradicted the original electrostatic arrangement, have we?
You might be able to shed light on this.
Thanks.
Hi,
I'm a little confused about what you were saying about "the opposite direction" and "contradiction". Could you explain that again please?
As for the electric field, we say it's zero inside the sphere - not because there is no electric field, but because all the field lines are pointing in all sorts of directions so that they end up canceling out to zero. So when you say that the electric field could be nonzero somewhere inside the sphere, I don't think that's right (but I could be wrong).
Then, if you were to add a charge (assuming positive) inside the sphere, my guess is that the charge would immediately go to the outer edge of the sphere so that the charge enclosed is still zero.
@@danthetutor2624
Hi thanks for your reply.
All I mean by implication in the "opposite direction" is:
1. If E=0 everywhere inside, then the integral of E over the surface is zero, then q_enclosed is zero (one direction of implication).
2. But, if q_enclosed is zero, this doesn't mean E everywhere inside has to be zero (implication in the opposite direction is false; the converse is not true essentially).
This is just from the mathematics: if the integrand E is zero, the integral (evaluated as q_enclosed) is zero but if the integral is zero, the integrand E doesn't have to be zero everywhere.
Regarding your second paragraph, I'm also talking about E being the net electric field at any point inside: I think we are on the same page about that. As you say, it is proposed that the electric field contributions from all the charges cancel at every point inside i.e. that E=0 everywhere inside. I'm not saying E is zero or E is not zero: I'm just saying we can't find whether E is or is not zero just from the fact that q_enclosed=0, as seems to have been done here.
Anyway, whatever about that, my main question has to do with how this E=0 is arrived at.
Books arrive at it by "proof by contradiction" (that's what I mean by contradiction above).
So it typically goes like:
1. Take an electrostatic arrangement of charge on a conducting solid sphere.
2. The net E field everywhere inside must be zero.
3. For If it wasn't zero at a point inside, we could place a charge at that point and obviously the charge would start moving.
4. This moving charge contradicts our electrostatic arrangement assumption above.
5. So we can't have a non-zero net E at any point inside.
My issue with this is step 3 seems to be a (slightly) different electrostatic arrangement of charge to step 1. We've changed the situation slightly by placing an extra charge inside. So this usual argument in textbooks seems flawed.
How I've tried to resolve it is by assuming that E fields (and E field lines) are continuous in free space. So if we had a non-zero E field at any point inside, there will an (attenuated) extra non-zero E field where the actually charges are themselves, moving them and changing the electrostatic arrangement. Which is a contradiction. So the net E field inside everywhere is zero.
I think I can accept that E fields are continuous in free space (by Gauss's law) so my last paragraph makes sense to me. The steps above (how it's presented in textbooks) doesn't quite make sense.
Can you comment on any of that.
Thanks
Ok, I see what you're saying. I hate trying to read textbook explanations. I also don't think I can add anything to what you've already said because that's about my understanding of it. If you're in school or university right now, I would probably ask one of your teachers.
@@danthetutor2624
Ah that's fine. That's for getting back. I'm always just trying to refine my understanding and check if people like yourself might know something that I'm just not getting. It can be frustrating, especially if it's probably something insignificant like just misinterpreting a poorly worded textbook explanation as you say. But cheers anyway. Appreciate your help.
If r=a?????????
You can just treat that as r>a. A gaussian drawn on the surface of the sphere would include all the charge and be no different from any other gaussian drawn further out ;)