I understand the case where the sphere is grounded. There you can set V=0 and you have clear boundary conditions for your Poisson equation. However I completely fail to understand the case where the sphere isn't grounded. What are the boundary conditions here? I thought the point was to take some image charges that yield the same boundary conditions for the Poisson equation but these conditions are nowhere mentioned. How would we know what the potential on the surface of the non-grounded conductor is in advance? Isn't this essential information that we need before we can even start using the method of images? I've breen trying to understand this for days and it's really driving me mad.
@@barnabasperenyi1785 I think I understand it better now. So we start with an uncharged spherical conductor. We bring a charge Q near it. The charge on the sphere redistributes itself and the surface is now at a constant potential V. We don't know the value of this potential yet, but it doesn't matter I think. In our images setup, we set the first charge q as before and then add a charge q' at the center that will make sure the spherical shell is at this potential V, which we still don't know exactly but whatever it is, there is some carge that gives it. With this setup, the potential and therefore the electric field must be equal in the original setup and the images setup, as long as we're outside the shell. This follows from the unicity of the solution to the Poisson equation with boundary conditions. Now we can use Gauss's Law. Take a Gaussian surface that encloses the conductor but not the charge Q. The flux through this surface must be zero because the sphere has no net charge. Now take the same Gaussian surface in the images setup with the three point charges. There is no conductor here but still the electric field is the same when we are outside the region. Since E is the same everywhere on the surface, the flux through this surface is the same and must also be zero. Since the flux through a closed surface depends only on the enclosed charge, the total charge inside the surface must be zero. Hence q'=-q. From this we would also be able to calculate the potential on the conducting sphere's surface.
If there's a net charge on the conductor, start with the solution for the non-grounded neutral sphere, but include yet another image charge at the centre of the sphere, with a charge equal to the total charge on the conductor.
@@mra8554 You're right, we can't place image charges in the region we're solving in - they'd introduce singularities that shouldn't be there. In fact, I don't think you'd need an extra image charge at all if you're solving inside the sphere. You can view the charge distribution on the sphere as a superposition of the original uniform charge density, plus an extra charge density induced by the point charge. The uniform component doesn't actually affect the field inside, which you can see from Gauss' law (or equivalently Newton's shell theorem), so the conclusion is that adding a nonzero surface charge to the conductor doesn't make any difference inside.
Well, not a total coincidence as we've chosen the charge and distance to be just right to make the potential zero on that sphere. But I suppose we are lucky in the sense that a spherical equipotential exists in the first place. It would be impossible to apply the method of images in the case of a cube - I can't imagine any arrangement of charges that would give a cube-shaped equipotential!
@@DrBenYelverton but the thing is when I computed th surface charge density I am getting a variable factor (1/a- cos theta)^(3/2) and I am not being able include the fact that one half has negative charge density and other has positive.... I have only been able to figure out the nature of variation
In the video we interpreted the charge outside as "real" and the charge inside as the image. But, we're free to interpret it the other way around, with the charge inside being "real" and the charge outside being the image. That means the solution for the case you described is the same as in the video. So for a charge Q at distance d, whether d > r or d < r, the image charge will be at a distance r²/d with a charge q = -(r/d)Q.
@@DrBenYelverton Wouldn’t the point charge inside the conducting sphere produce electric field and breaks the electrostatic equilibrium of the sphere? Can we still solve the problem using the “inverse” method of images you’ve mentioned above? I tried to imagine the case that the point charge is in a cavity that covered the charge so that the sphere will still maintain electrostatic equilibrium. But it seems complicated how method of image will be used in this case😢(I can’t define the boundaries due to the cavity). Is there any misunderstanding and what approach can I take?And great video professor! I really enjoy your explanation.
I have to calculate point charge, near infinite grounded tube with a set radius. Is it even possible without doing it numericaly? I think it would look similary to infinite line of charge near a tube, but the charge density would change along the tube. Is it doable?
It's not possible to solve exactly - it seems that approximate solutions do exist but the maths is very involved, even in the symmetrical case where the charge is at the centre of the cylinder! See e.g. arxiv.org/abs/2001.10651
Another very good e.m. video! Thanks!
I understand the case where the sphere is grounded. There you can set V=0 and you have clear boundary conditions for your Poisson equation.
However I completely fail to understand the case where the sphere isn't grounded. What are the boundary conditions here? I thought the point was to take some image charges that yield the same boundary conditions for the Poisson equation but these conditions are nowhere mentioned. How would we know what the potential on the surface of the non-grounded conductor is in advance? Isn't this essential information that we need before we can even start using the method of images?
I've breen trying to understand this for days and it's really driving me mad.
looking for the same answer here
@@barnabasperenyi1785 I think I understand it better now. So we start with an uncharged spherical conductor. We bring a charge Q near it. The charge on the sphere redistributes itself and the surface is now at a constant potential V. We don't know the value of this potential yet, but it doesn't matter I think.
In our images setup, we set the first charge q as before and then add a charge q' at the center that will make sure the spherical shell is at this potential V, which we still don't know exactly but whatever it is, there is some carge that gives it.
With this setup, the potential and therefore the electric field must be equal in the original setup and the images setup, as long as we're outside the shell. This follows from the unicity of the solution to the Poisson equation with boundary conditions.
Now we can use Gauss's Law. Take a Gaussian surface that encloses the conductor but not the charge Q. The flux through this surface must be zero because the sphere has no net charge. Now take the same Gaussian surface in the images setup with the three point charges. There is no conductor here but still the electric field is the same when we are outside the region. Since E is the same everywhere on the surface, the flux through this surface is the same and must also be zero. Since the flux through a closed surface depends only on the enclosed charge, the total charge inside the surface must be zero. Hence q'=-q. From this we would also be able to calculate the potential on the conducting sphere's surface.
If the conductor had some non zero surface charge density, then what approach should I take?
If there's a net charge on the conductor, start with the solution for the non-grounded neutral sphere, but include yet another image charge at the centre of the sphere, with a charge equal to the total charge on the conductor.
@@DrBenYelverton ok thankyou🙏
but if your original source is inside the sphere, then an image charge cannot be placed in the center right?
@@DrBenYelverton
@@mra8554 You're right, we can't place image charges in the region we're solving in - they'd introduce singularities that shouldn't be there. In fact, I don't think you'd need an extra image charge at all if you're solving inside the sphere. You can view the charge distribution on the sphere as a superposition of the original uniform charge density, plus an extra charge density induced by the point charge. The uniform component doesn't actually affect the field inside, which you can see from Gauss' law (or equivalently Newton's shell theorem), so the conclusion is that adding a nonzero surface charge to the conductor doesn't make any difference inside.
is it coincidence that the plane of potential zero is a sphere? If I had for example a grounded cube, how would that translate to the problem?
Well, not a total coincidence as we've chosen the charge and distance to be just right to make the potential zero on that sphere. But I suppose we are lucky in the sense that a spherical equipotential exists in the first place. It would be impossible to apply the method of images in the case of a cube - I can't imagine any arrangement of charges that would give a cube-shaped equipotential!
where is the distance taken from
From the origin
if it was just a non grounded neutral conducting sphere and a point charge, how to approach the problem?
This is discussed at the end of the video, from around 15:00 onwards.
@@DrBenYelverton but the thing is when I computed th surface charge density I am getting a variable factor (1/a- cos theta)^(3/2) and I am not being able include the fact that one half has negative charge density and other has positive.... I have only been able to figure out the nature of variation
If point charge inside a conducting sphere, then what approach should I take?
In the video we interpreted the charge outside as "real" and the charge inside as the image. But, we're free to interpret it the other way around, with the charge inside being "real" and the charge outside being the image. That means the solution for the case you described is the same as in the video. So for a charge Q at distance d, whether d > r or d < r, the image charge will be at a distance r²/d with a charge q = -(r/d)Q.
@@DrBenYelverton Wouldn’t the point charge inside the conducting sphere produce electric field and breaks the electrostatic equilibrium of the sphere? Can we still solve the problem using the “inverse” method of images you’ve mentioned above? I tried to imagine the case that the point charge is in a cavity that covered the charge so that the sphere will still maintain electrostatic equilibrium. But it seems complicated how method of image will be used in this case😢(I can’t define the boundaries due to the cavity). Is there any misunderstanding and what approach can I take?And great video professor! I really enjoy your explanation.
I have to calculate point charge, near infinite grounded tube with a set radius. Is it even possible without doing it numericaly? I think it would look similary to infinite line of charge near a tube, but the charge density would change along the tube. Is it doable?
It's not possible to solve exactly - it seems that approximate solutions do exist but the maths is very involved, even in the symmetrical case where the charge is at the centre of the cylinder! See e.g. arxiv.org/abs/2001.10651
@@DrBenYelverton thank you very much!