Force between two spheres connected by wire and battery

[Saitama]

Yes.

Since the sphere is neutral, the charge introduced should be of same magnitude as image charge. But I am unsure about the position of this new charge. If I place it at centre of right sphere, then I can subtract the repulsive force from F to get the answer but I don't see the reason behind placing the charge at centre of right sphere.

 

[TSny]

I don't see the reason behind placing the charge at centre of right sphere.

You need to keep the surface of the sphere an equipotential surface.

 

[Saitama]

You need to keep the surface of the sphere an equipotential surface.

Okay, this could sound dumb but why do we need to keep the surface equipotential?

 

[Saitama]

Think about it.

I honestly don't see it. I tried writing down the equations to see if placing the extra charge at centre makes the surface of sphere equipotential.

The magnitude of extra charge is same as the induced charge. If this charge is placed at the centre of sphere, the potential due to it on the surface of sphere is $kq/R$ where $q=QR/d$.

If I consider a point on the sphere and find the net potential due to these two charges, it comes out to be different for different positions of the point under consideration on the surface of sphere.

 

[TSny]

If I consider a point on the sphere and find the net potential due to these two charges, it comes out to be different for different positions of the point under consideration on the surface of sphere.

That's right. But there's a third charge (the charge way over to the left on the left sphere). The surface of the sphere on the right must be an equipotential surface under the influence of all the charges in the entire system.

The whole purpose of the first image charge is to create a net potential of zero on the surface of the right sphere when you consider just the image charge and the charge on the left sphere. When you add the extra image charge to make the right sphere neutral, it must be done in a way so that the surface of the right sphere is an equipotential surface for all of the charges in the system.

 

[SammyS]

I honestly don't see it. I tried writing down the equations to see if placing the extra charge at centre makes the surface of sphere equipotential.

The magnitude of extra charge is same as the induced charge. If this charge is placed at the centre of sphere, the potential due to it on the surface of sphere is $kq/R$ where $q=QR/d$.

If I consider a point on the sphere and find the net potential due to these two charges, it comes out to be different for different positions of the point under consideration on the surface of sphere.

The spheres are conductors.

 

[Saitama]

The whole purpose of the first image charge is to create a net potential of zero on the surface of the right sphere when you consider just the image charge and the charge on the left sphere. When you add the extra image charge to make the right sphere neutral, it must be done in a way so that the surface of the right sphere is an equipotential surface for all of the charges in the system.

I think I get it now, thanks TSny!

But I have a doubt about applying the method of images in the present case. The derivation of image charge I have seen online or in books grounds the sphere or set the sphere at zero potential. The problem statement doesn't mention that the sphere is at zero potential, then why is it legit to use method of images?

 

[TSny]

The reason why the books usually ground the sphere is to make it easier! Grounding the sphere puts the sphere at zero potential. Then you only need one image charge inside the sphere to make the total potential zero on the sphere. If you want to raise the potential of the surface to some non-zero value, then you can add another point charge at the center of the sphere. The surface of the sphere will still be an equipotential surface, but it will have a non-zero value.

In part (b), you need to have zero net charge for the right sphere. This tells you how much charge to add at the center. The surface of the right sphere will be an equipotential surface but the potential will not be zero.

 

[SammyS]

To continue on what TSny said:

The image charge and the charge at the center being equal in magnitude but opposite in sign constitutes a dipole. The sphere on the right acts like a dipole in this case, which is induced by the electric field produced by other sphere .

 

[Saitama]

The reason why the books usually ground the sphere is to make it easier! Grounding the sphere puts the sphere at zero potential. Then you only need one image charge inside the sphere to make the total potential zero on the sphere. If you want to raise the potential of the surface to some non-zero value, then you can add another point charge at the center of the sphere. The surface of the sphere will still be an equipotential surface, but it will have a non-zero value.

In part (b), you need to have zero net charge for the right sphere. This tells you how much charge to add at the center. The surface of the right sphere will be an equipotential surface but the potential will not be zero.

To continue on what TSny said:

The image charge and the charge at the center being equal in magnitude but opposite in sign constitutes a dipole. The sphere on the right acts like a dipole in this case, which is induced by the electric field produced by other sphere .

Thanks a lot both of you! :)

Proceeding with part c), I think its the same situation as b) except that the right sphere is set at zero potential. The given answer is similar to the expression for F I obtained but it seems like I have to make the second term disappear. How should I do that?

 

[TSny]

Because d >> r, you can keep only the lowest order term in r/d.

 

[Saitama]

Because d >> r, you can keep only the lowest order term in r/d.

I don't get this. Reasoning like this, I can say that answer is zero for case b).

 

[TSny]

Sorry, I should have said keep only the lowest order non-zero term.

 

[Saitama]

Sorry, I should have said keep only the lowest order non-zero term.

Okay, thanks a lot TSny!

 

[SammyS]

I don't get this. Reasoning like this, I can say that answer is zero for case b).

Sorry, I should have said keep only the lowest order non-zero term.

Comparing answers for (a), (b), and (c), we see that as d becomes large compared to r, the answer to (c) gets small compared to the answer to (a) . The the answer to (b) gets extremely small compared to the answers to (a) or (c) . (So to a fairly reasonable approximation, rudeman was correct about this.)

All of these solutions are approximations correct to the lowest order non-zero term.

Take part (a) for instance:

We replace each of the spheres with charge Q and -Q, respectively, each placed at the location of the spheres' centers. This does give two spherical equi-potential surfaces of radius, r, which have a potential difference of U. However, the centers of the spheres are slightly more than d apart.

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