Force between two spheres connected by wire and battery

[Saitama]

Homework Statement

There are two alike initially uncharged spheres of radius r at a distance of d>>r, as shown in the figure.

​

a) Switch $K_2$ is closed. What is the force exerted between the two spheres?

b) What would be the force between the two spheres, if in the previous experiment instead of closing switch $K_2$, switch $K_1$ was closed?

c) Determine the force exerted between the spheres if both switches are closed.

Homework Equations

The Attempt at a Solution

In part a), if the battery sends a charge of $Q$ to the left sphere, then it must send a charge of -Q to the right sphere. Equating the potential difference gives:
$$\frac{2kQ}{R}=U \Rightarrow Q=\frac{UR}{2k}$$
Hence, the magnitude of force between the spheres is:
$$F=\frac{kQ^2}{d^2}=\frac{kU^2R^2}{4k^2d^2}$$
The force is attractive.

Is this correct?

In part b), I really have no idea. How do I deal with this case?

Any help is appreciated. Thanks!

 

[rude man]

part (a): right.

part (b): can any charge ever accumulate on either sphere?

part (c): what's the potential difference between the spheres now?

 

[Saitama]

part (b): can any charge ever accumulate on either sphere?

I am not sure, it feels as if the charge won't ever reach to the right sphere given the connection is grounded in between but I think I am wrong.

 

[rude man]

I am not sure, it feels as if the charge won't ever reach to the right sphere given the connection is grounded in between but I think I am wrong.

In order for charges to accumulate on the spheres, don't you need current to flow between them at some point, in this case displacement current? Can that happen when switch K2 is never closed?

 

[Saitama]

In order for charges to accumulate on the spheres, don't you need current to flow between them at some point, in this case displacement current? Can that happen when switch K2 is never closed?

I am still very much confused. Also, I am not satisfied with what I did in part a). Even if I close the switch $K_2$ how the charge will ever reach the spheres (i.e a current should flow) when the circuit isn't even complete?

 

[rude man]

I am still very much confused. Also, I am not satisfied with what I did in part a). Even if I close the switch $K_2$ how the charge will ever reach the spheres (i.e a current should flow) when the circuit isn't even complete?

The circuit IS complete. It's via the capacitance of the two spheres in mutual proximity.

Are you at all electronics-oriented? You can answer all three questions easily by thinking of what happens in a capacitor connected by one lead by the + terminal of a battery and with the various connections defined in the three parts of your problem.

 

[Saitama]

The circuit IS complete. It's via the capacitance of the two spheres in mutual proximity.

Can I say that the two spheres represent two separate spherical capacitors? For both the spherical capacitors, the one at infinity is grounded so the circuit for part a) can be drawn as shown in attachment. Is it correct?

Are you at all electronics-oriented?

I am not sure what do you mean by this.

 

[rude man]

Can I say that the two spheres represent two separate spherical capacitors? For both the spherical capacitors, the one at infinity is grounded so the circuit for part a) can be drawn as shown in attachment. Is it correct?


I am not sure what do you mean by this.

No, the two spheres make up ONE capacitor. Each sphere is one "plate" of the capacitor.

 

[Saitama]

No, the two spheres make up ONE capacitor. Each sphere is one "plate" of the capacitor.

So in part b), there would be no force of attraction? For part b), only one plate is connected to capacitor so the circuit is open, no charge to the capacitor, right?

What about c)? Can I redraw it as shown in the attachment?

 

[rude man]

So in part b), there would be no force of attraction? For part b), only one plate is connected to capacitor so the circuit is open, no charge to the capacitor, right?

What about c)? Can I redraw it as shown in the attachment?

Right on all counts!

 

[Saitama]

Right on all counts!

Thanks rude man for all the help so far!

Isn't part c) the same as part a) except the grounding connection? How would it affect the charges on the plates of capacitor?

 

[rude man]

Thanks rude man for all the help so far!

Isn't part c) the same as part a) except the grounding connection? How would it affect the charges on the plates of capacitor?

What's the potential between the spheres?

 

[Saitama]

What's the potential between the spheres?

The potential difference is U? Does that mean the force would be same as in part a)?

 

[rude man]

The potential difference is U? Does that mean the force would be same as in part a)?

Are you familiar with the First Uniqueness Theorem in electrostatics?

 

[Saitama]

Are you familiar with the First Uniqueness Theorem in electrostatics?

No. :(

 

[rude man]

No. :(

Well, is the potential difference between the spheres any different than in part (a)?

Hint: knowing V you can assume charge +Q and -Q on the spheres, use Gauss's theorem to determine the E fields generated by both spheres, integrate the force required on a unit test charge in going from the - sphere to the + sphere, equate to V, then that'll give you Q and of course then it's just the Coulomb calculation.

 

[Saitama]

Well, is the potential difference between the spheres any different than in part (a)?

No, its the same as in a).

Hint: knowing V you can assume charge +Q and -Q on the spheres, use Gauss's theorem to determine the E fields generated by both spheres, integrate the force required on a unit test charge in going from the - sphere to the + sphere, equate to V, then that'll give you Q and of course then it's just the Coulomb calculation.

Before doing the math, isn't the condition in c) is same as a)? The spheres with charge Q and -Q are at potential kQ/R and -kQ/R respectively, so the potential difference, 2kQ/R, must be equal to U. This would give me the same answer as a).

 

[rude man]

No, its the same as in a).

Before doing the math, isn't the condition in c) is same as a)? The spheres with charge Q and -Q are at potential kQ/R and -kQ/R respectively, so the potential difference, 2kQ/R, must be equal to U. This would give me the same answer as a).

That's correct, but since you didn't know about the uniqueness theorem I thought you might want to confirm your conclusion a second way.

 

[Saitama]

That's correct, but since you didn't know about the uniqueness theorem I thought you might want to confirm your conclusion a second way.

Thanks rude man once again! :)

I have seen the uniqueness theorem in books by Purcell and Griffiths but I used those books only for a short period of time so I couldn't gain much from them. Moreover, I think it involves a lot of math which isn't taught at a high school level.

 

[rude man]

Thanks rude man once again! :)

I have seen the uniqueness theorem in books by Purcell and Griffiths but I used those books only for a short period of time so I couldn't gain much from them. Moreover, I think it involves a lot of math which isn't taught at a high school level.

You're doing pretty impressive work for a high school student! Keep it up!

 

[Saitama]

Sorry for the bump but I recently got the answers to this problem and it looks as if we got only the first one correct.

The correct answers as follows:

For a): $\frac{U^2}{4k}\left(\frac{r}{d}\right)^2$

For b): $\frac{2U^2}{k}\left(\frac{r}{d}\right)^5$

For c): $\frac{U^2}{k}\left(\frac{r}{d}\right)^3$

Rude man, can you please provide some further assistance? That would be greatly appreciated. Thanks!

 

[rude man]

Sorry for the bump but I recently got the answers to this problem and it looks as if we got only the first one correct.

The correct answers as follows:

For a): $\frac{U^2}{4k}\left(\frac{r}{d}\right)^2$

For b): $\frac{2U^2}{k}\left(\frac{r}{d}\right)^5$

For c): $\frac{U^2}{k}\left(\frac{r}{d}\right)^3$

Rude man, can you please provide some further assistance? That would be greatly appreciated. Thanks!

Pranav, I am at a loss. I can't see how those answers are correct (except (a). Maybe the sequence of switch-throwing is different than I thought. Maybe someone else will jump in.

 

[SammyS]

Sorry for the bump but I recently got the answers to this problem and it looks as if we got only the first one correct.

The correct answers as follows:

For a): $\frac{U^2}{4k}\left(\frac{r}{d}\right)^2$

For b): $\frac{2U^2}{k}\left(\frac{r}{d}\right)^5$

For c): $\frac{U^2}{k}\left(\frac{r}{d}\right)^3$

Rude man, can you please provide some further assistance? That would be greatly appreciated. Thanks!

Pranav-Arora,

I considered jumping in earlier, when rudeman implied that no charge could accumulate on either sphere in part b, but I didn't have any good approach to solving either part b or part c .

I have not solved these, but I do have some ideas.

Part b: I think this is the tougher one.

Charge will accumulate on the left hand sphere. It's potential will be at U w.r.t ground, which I suppose is infinitely far away.

Then the potential of the right hand sphere is the same as for any point a distance of d from the left hand sphere.

A small amount of excess negative charge will accumulate on one side of the isolated sphere. Positive on the other side. (It's neutral.)

I think this effect might possibly be handled by the method of image charges.​

Part c:

The left hand sphere is at a potential U w.r.t. a point that's distance d away.

The surface of the grounded sphere is at zero potential.

Again use an image charge. What charge is necessary to make the surface of the grounded sphere to be at zero potential?​

 

[rude man]

@Pranav, I confess I never considered the Earth as part of the picture. I will leave it to SammyS and others to carry the ball from here on. Sorry I seem to have let you down this time.

 

[Saitama]

Hi SammyS!

Part b: I think this is the tougher one.

Charge will accumulate on the left hand sphere. It's potential will be at U w.r.t ground, which I suppose is infinitely far away.

Then the potential of the right hand sphere is the same as for any point a distance of d from the left hand sphere.

A small amount of excess negative charge will accumulate on one side of the isolated sphere. Positive on the other side. (It's neutral.)

I think this effect might possibly be handled by the method of image charges.​

I am not sure but I need the charge on the left sphere to find the image charge, how do I find it?

Its at a potential U wrt ground so I suppose the charge on it is $Q=UR/k$? Do I have to find the image charge of left sphere in the right sphere? That looks difficult, do I assume the left sphere as a point charge while finding the image?

@Pranav, I confess I never considered the Earth as part of the picture. I will leave it to SammyS and others to carry the ball from here on. Sorry I seem to have let you down this time.

Please don't be sorry, I am glad to know you tried to help. Everybody makes mistakes. :)

 

[SammyS]

Hi SammyS!

I am not sure but I need the charge on the left sphere to find the image charge, how do I find it?

It's at a potential U wrt ground so I suppose the charge on it is $Q=UR/k$? Do I have to find the image charge of left sphere in the right sphere? That looks difficult, do I assume the left sphere as a point charge while finding the image?

Please don't be sorry, I am glad to know you tried to help. Everybody makes mistakes. :)

Sorry for the tardiness of this reply.

I hoped someone else might jump in.

$Q=UR/k$ looks good to me -- at least for part b. I'm still debating part c .

To find the image charge of left sphere in the right sphere: Yes, treat the left as being a point charge. They're far apart.

 

[Saitama]

Sorry for the tardiness of this reply.

I hoped someone else might jump in.

$Q=UR/k$ looks good to me -- at least for part b. I'm still debating part c .

To find the image charge of left sphere in the right sphere: Yes, treat the left as being a point charge. They're far apart.

I think I am reaching close to the answer of part b) but I am stuck at the last step. Here's what I did:

The image charge is of magnitude $QR/d$ and is at distance of $R^2/d$ from the centre of right sphere. The distance between the image charge and the left sphere is given by $d-R^2/d$. Hence, the force of attraction is:
$$F=\frac{kQ^2R}{d\left(d-\frac{R^2}{d}\right)^2}=\frac{U^2R^3}{kd^3}\left(1-\frac{R^2}{d^2}\right)^{-2}\approx \frac{U^2R^3}{kd^3} + \frac{2U^2}{k}\left(\frac{R}{d}\right)^5$$
To reach the given answer, I somehow need to make the first term disappear but I don't see how.

 

[TSny]

In part (b) the right sphere is isolated. So, what must the total charge be for the right sphere?

The answer you found for F might be useful for part (c), where the right sphere is grounded!

 

[Saitama]

HI TSny! :)

In part (b) the right sphere is isolated. So, what must the total charge be for the right sphere?

Zero? Does this mean I need to introduce an extra charge to make it neutral again?

The answer you found for F might be useful for part (c), where the right sphere is grounded!

 

[TSny]

Zero? Does this mean I need to introduce an extra charge to make it neutral again?

Yes.

 

[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? :uhh:

 

[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.