Charge rearrangement on conducting spheres

  • #1
261
26

Homework Statement:

A conductive sphere of radius ## R ## carries a free charge ## Q_0 ##. A second conductive sphere is neutral. The spheres are brought into contact. Find the charge on each sphere when they are separated again.

Relevant Equations:

The formula giving the electric potential outside a charged isolated sphere
Hi,
I think this problem is solved in exactly as a similar problem where the two spheres are very far apart and connected by a very long thin conducting wire. I'm trying to explain this in words, since LaTeX does not seem to work any more (for some reason LaTeX syntax is not replaced by maths in the preview).

So my reasoning is the following.
  1. In the "experiment" involved in the problem the two spheres actually touch. The charged sphere will loose some of its charge, which will transfer onto the previously neutral sphere.
  2. While touching, the two spheres form a singular conductor. The charge stays on the surface of this conductor and rearrange so that the electric field inside the conductor vanishes.
  3. Since all the charge has the same sign, it tends to spread. It won't spread uniformly, though. I expect that there is basically no charge in the vicinity of the contact point.
  4. When the spheres are separated again, the amount of charge on each of them stays the same. I see no reason for some charge migrating from one sphere to the other.
  5. The key step is now to assume that the amount of charge on each sphere is the same as in a different "experiment". where the spheres are infinitely apart and are connected through a conducting wire.
  6. Since there is an "escape route", some charge leaves the originally charged sphere and moves onto the originally neutral one. Another key assumption is that the wire is so thin that the amount of charge that remains on it is negligible.
  7. Since the spheres are connected, they form a singular conductor, and are therefore at the same potential.
  8. Since the spheres are infinitely apart, the charge onto their surfaces spreads uniformly. Therefore electric potential on their surface is the same as that of a suitable point charge placed at the center of the sphere. Such potential is proportional to the point charge and inversely proportional to the surface radius.
  9. Equating the potentials one obtains that the ratio of the charges on the spheres is the same as the ratio of their radii. The final charges are obtained from this condition, along with the fact that their sum should give the original charge.
I think this works, but I'd like to have a stronger argument for the assumption in 5.
Also, I wonder whether there is a simpler overall argument. A crucial part of my argument is that the potential on the surface of each sphere is the same when they are infinitely apart. But what is the reason of that, if there is no connecting wire?

Of course the potential on the surface of the spheres (and inside them) is the same when they touch. But in that case (and, in general, when they are not infinitely apart) the charge distribution on their surfaces is far from uniform. Also, I'm not sure the surface potentials stay the same when the sphere are separated.

Thanks for any insight
 
  • Like
Likes etotheipi and Delta2

Answers and Replies

  • #2
rude man
Homework Helper
Insights Author
Gold Member
7,856
792
What's the radius of the second sphere?
 
  • #3
etotheipi
Gold Member
2019 Award
2,943
1,899
If I understand correctly, the experiment you are doing involves bringing the spheres together so that they touch, and then bringing them back to infinity (with no wire involved).

This is as opposed to the other example you mention, where you would bring them together so that they touch, connect a conducting wire between them so that they are constrained to be at equal potential until you've brought them back to infinity, and then cut the wire.

I think it is a good question. If the spheres have different radii, I don't think it is obvious that the line integral of the electric field from infinity either sphere at any given point during the separation process has to be the same for both spheres! Even though this seems like it would make sense.

It seems to only be obvious when the spheres have equal radii, because of symmetry considerations (i.e. suppose one ended up at a higher potential than the other; we would have no way of "choosing" such a favourite, so they must be at the same potential).

But I am also interested to see if anyone has a good answer for the case where the radii are not equal!
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,861
6,129
Two charged conducting spheres in proximity can be handled by an iterative version of the method of images, but it is not for the faint-hearted.
https://arxiv.org/pdf/0906.1617.pdf addresses two noncontacting spheres at unequal potentials. Not sure how readily it can be adapted to contacting spheres.
 
  • Informative
Likes etotheipi
  • #5
rude man
Homework Helper
Insights Author
Gold Member
7,856
792
Yeah, the problem is simple if the spheres stay connected 'till they're infinity apart. Much less so otherwise. @haruspex, that Maxwell paper you once sent me - doesn't that address the latter contingency?
 
  • Like
Likes etotheipi
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,861
6,129
Yeah, the problem is simple if the spheres stay connected 'till they're infinity apart. Much less so otherwise. @haruspex, that Maxwell paper you once sent me - doesn't that address the latter contingency?
I couldn't find the paper that led me to the Maxwell one. Did just find this: http://www.hep.princeton.edu/~mcdonald/examples/twospheres.pdf, which does discuss equipotent spheres and links to a copy of a Maxwell paper, but a paywalled one. Can probably find free access copies with a suitable search.

Edit:
This looks like the paper I found previously https://royalsocietypublishing.org/doi/10.1098/rspa.2012.0133.
It deals specifically with the case of touching spheres.

A key claim of the paper is to show that conducting spheres of like charge will attract if close enough, except for the case where their charges are already in the ratio they would have when in contact. In that special case they will repel.
Now, that doesn't make sense to me. Surely the force between them, at a given distance, varies continuously with the charge ratio. How can it jump suddenly from attraction to repulsion when the charge ratio hits the magic value?
But I have not attempted to find an error in the paper.
 
Last edited:
  • Like
Likes etotheipi
  • #7
Delta2
Homework Helper
Insights Author
Gold Member
3,235
1,196
Though this problem doesn't seem to have an exact analytical solution, i believe that up to a very good degree of approximation the solution presented in the OP(that the ratio of charges equals the ratio of the radius of the two spheres) is the solution of the problem.
First of all we gonna work in the quasi-static approximation where the E-field is conservative at all times. This means that the line integral of the E-field is the same between two end points regardless of the path between the two points. So lets consider a point A on the surface of sphere with radius ##r_1## and a point B on the surface of the sphere with radius ##r_2##. It will be
$$V_A=\int_{\mathbf{r_A}}^{\infty}\mathbf{E}\cdot d\mathbf{l}=\int_{\mathbf{r_A}}^{\mathbf{r_B}}\mathbf{E}\cdot d\mathbf{l}+\int_{\mathbf{r_B}}^{\infty}\mathbf{E}\cdot d\mathbf{l}=\int_{\mathbf{r_A}}^{\mathbf{r_B}}\mathbf{E}\cdot d\mathbf{l}+V_B$$
Now if we could only prove the ##\int_{\mathbf{r_A}}^{\mathbf{r_B}}\mathbf{E}\cdot d\mathbf{l}=0## or equivalently that $$\int_C\mathbf{E}\cdot d\mathbf{l}=0 \text { (1)}$$ where ##C## is the straight path that connects the centers of the two spheres, then we would have that ##V_A=V_B##
I think we can prove (1) as long as we assume that the charges on the two spheres are such that $$\frac{Q_1}{r_1}=\frac{Q_2}{r_2}\text{ (2)}$$ and the mild assumption that the electric field in the straight path that connects the centers of the two spheres is that of two points charges ##Q_1## and ##Q_2## that lie at the centers of the two spheres respectively, (and also that the electric field is zero in portion of the straight path that corresponds to the interior of the two spheres).
So under the charge distribution of (2) and this mild assumption we ll have that ##V_A=V_B## always as we move the spheres away from each other, without the need for them to be connected by a thin wire.
 
  • Like
Likes etotheipi
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,861
6,129
Though this problem doesn't seem to have an exact analytical solution, i believe that up to a very good degree of approximation the solution presented in the OP(that the ratio of charges equals the ratio of the radius of the two spheres) is the solution of the problem.
First of all we gonna work in the quasi-static approximation where the E-field is conservative at all times. This means that the line integral of the E-field is the same between two end points regardless of the path between the two points. So lets consider a point A on the surface of sphere with radius ##r_1## and a point B on the surface of the sphere with radius ##r_2##. It will be
$$V_A=\int_{\mathbf{r_A}}^{\infty}\mathbf{E}\cdot d\mathbf{l}=\int_{\mathbf{r_A}}^{\mathbf{r_B}}\mathbf{E}\cdot d\mathbf{l}+\int_{\mathbf{r_B}}^{\infty}\mathbf{E}\cdot d\mathbf{l}=\int_{\mathbf{r_A}}^{\mathbf{r_B}}\mathbf{E}\cdot d\mathbf{l}+V_B$$
Now if we could only prove the ##\int_{\mathbf{r_A}}^{\mathbf{r_B}}\mathbf{E}\cdot d\mathbf{l}=0## or equivalently that $$\int_C\mathbf{E}\cdot d\mathbf{l}=0 \text { (1)}$$ where ##C## is the straight path that connects the centers of the two spheres, then we would have that ##V_A=V_B##
I think we can prove (1) as long as we assume that the charges on the two spheres are such that $$\frac{Q_1}{r_1}=\frac{Q_2}{r_2}\text{ (2)}$$ and the mild assumption that the electric field in the straight path that connects the centers of the two spheres is that of two points charges ##Q_1## and ##Q_2## that lie at the centers of the two spheres respectively, (and also that the electric field is zero in portion of the straight path that corresponds to the interior of the two spheres).
So under the charge distribution of (2) and this mild assumption we ll have that ##V_A=V_B## always as we move the spheres away from each other, without the need for them to be connected by a thin wire.
The Royal Society link I gave in my edit to post #6 states that a good approximation to the charge ratio of unequal conducting spheres in contact is ##(\frac ba)^2(\frac{\pi^2}6)^\frac{a−b}{a+b}##;
 
  • Like
Likes etotheipi
  • #9
Delta2
Homework Helper
Insights Author
Gold Member
3,235
1,196
The Royal Society link I gave in my edit to post #6 states that a good approximation to the charge ratio of unequal conducting spheres in contact is ##(\frac ba)^2(\frac{\pi^2}6)^\frac{a−b}{a+b}##;
They might be right, it turns out when I tried to do the math, i couldnt prove (1) given (2) and the "mild" assumption. (1) holds only exactly when the spheres are of equal radius, and it holds as a vague approximation when the spheres are of unequal radius.
 
  • #10
261
26
What's the radius of the second sphere?
Hi rude man,
you're right, I forgot to specify that. In general the second sphere has a radius ## r \neq R ##.
Of course if it was ## r=R ## it would be reasonable to conclude that the two spheres equally share the initial charge. I was planning to include these considerations in my OP, but in the end it slipped my mind.
Sorry about that.
 
  • #11
rude man
Homework Helper
Insights Author
Gold Member
7,856
792
Hi rude man,
you're right, I forgot to specify that. In general the second sphere has a radius ## r \neq R ##.
Of course if it was ## r=R ## it would be reasonable to conclude that the two spheres equally share the initial charge. I was planning to include these considerations in my OP, but in the end it slipped my mind.
Sorry about that.
Thass' alright! :smile:

@haruspex, I think I retained the Maxwell (or Maxwell-based) paper you sent , will look for it soon.

EDIT: If it's the Royal Soc. paper you cite I won't bother.
 
Last edited:
  • #12
261
26
[..]
I think we can prove (1) as long as we assume that the charges on the two spheres are such that $$\frac{Q_1}{r_1}=\frac{Q_2}{r_2}\text{ (2)}$$ and the mild assumption that the electric field in the straight path that connects the centers of the two spheres is that of two points charges ##Q_1## and ##Q_2## that lie at the centers of the two spheres respectively, (and also that the electric field is zero in portion of the straight path that corresponds to the interior of the two spheres).
So under the charge distribution of (2) and this mild assumption we ll have that ##V_A=V_B## always as we move the spheres away from each other, without the need for them to be connected by a thin wire.
Hi Delta2,
if I'm getting it right, you are using the fact that the potential "outside" a charged sphere is
$$ V = k \frac{Q}{r}\quad\text{(1)}$$
i.e. the same as a point charge ##Q## at the center of the sphere. But this is rigorously true only if the charge distribution on the surface of the sphere is uniform, right?
In general this is not the case, since che charges move freely throughout a conductor. The charge on the other sphere would disrupt the uniformity of the distribution, unless the spheres are very very far apart.

etotheipi said:
If I understand correctly, the experiment you are doing involves bringing the spheres together so that they touch, and then bringing them back to infinity (with no wire involved).
Yes, but bringing them back to infinity is just a trick to solve the problem.
The problem asks to find the final charge on each sphere. When the spheres are separated after touching the charge they carry stays the same, irrespective of the distance between the spheres. Bringing them very far apart results in such charges spreading uniformly on their surfaces, which makes the problem more symmetric, allowing e.g. to use equation (1).

The only way I can think of solving this problem without resort to ugly looking calculations is stil through the "thought experiment" involving the thin wire.

The key point of my reasoning is that the charge rearranges in the same way whether the spheres are brought in contact or connected through a thin wire.
We perhaps could think that when they are brought apart after touching, a very thin conducting wire keeps connecting them. This wire extends indefinitely as the distance between the spheres increases.
I see no reason for the charges on a sphere migrating to the other sphere during this process.
When the spheres are very far the problem is simple, because the potential on their surface is the same, and described by eq (1).
We can then think of severing the wire.

Perhaps another way of thinking about this is assuming that initially the spheres are separated but carry the same charges that would result from the situation where they are connected by a very long and thin wire.
What would happen if they are brought into contact. Would the charge ratio change?
 
Last edited:
  • #13
Delta2
Homework Helper
Insights Author
Gold Member
3,235
1,196
Hi Delta2,
if I'm getting it right, you are using the fact that the potential "outside" a charged sphere is
$$ V = k \frac{Q}{r}\quad\text{(1)}$$
i.e. the same as a point charge ##Q## at the center of the sphere. But this is rigorously true only if the charge distribution on the surface of the sphere is uniform, right?
In general this is not the case, since che charge move freely throughout a conductor. The charge on the other sphere would disrupt the uniformity of the distribution, unless the spheres are very very far apart.
Yes you are absolutely right here at your comments. I tried to use this as a simplifying assumption in order to prove that the potential of the two spheres remains approximately equal even after we move them apart, but i couldn't prove it. My conclusion is that we cant prove that the two experiments outlined in the OP are equivalent in terms of the resulting charge ratio. They are approximately equivalent if the two spheres have approximately the same radius. That's all i can say. @haruspex paper link at post #6 claims that when we bring two spheres in contact then the charge ratio is approximately $$\left ( \frac{r_1}{r_2}\right )^2\left (\frac{\pi^2}{6}\right )^\frac{r_1-r_2}{r_1+r_2}$$
which is quite different than ##\frac{r_1}{r_2}## . The two quantities are approximately the same if we take ##r_1\approx r_2##.
 
  • Like
Likes etotheipi
  • #14
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,861
6,129
I see no reason for the charges on a sphere migrating to the other sphere during this process.
There is one. When close, the potential on each sphere is altered by its proximity to the other sphere. The effect will be greater for the smaller sphere.
According to the approximation in post #6 (which is reckoned to be good both when the spheres are similar in radius and when very different) and taking the case where the ratio of the radii is 1+x, small x, the ratio of the charges approximates 1+3x/2.
 
  • Like
Likes etotheipi
  • #15
261
26
Uhm my book (the Italian version of "Physics", by Cutnell and Johnson) has examples and exercises where charged conductive spheres are "brought into contact" and share their charge. The final charges are always calculated assuming

$$ \frac{q_1}{q_2} = \frac{r_1}{r_2} $$

At this point this seems to be correct only if "brought into contact" means "connected via a very thin conductive wire while infinitely far from each other".
I guess that the textbook should give a bit more detail on what it means by "brought into contact", because the above situation is definitely not what one pictures in his/her mind.

This seems to be a problem of the Italian version of the book only. I skimmed through its original version, and could not find the examples and exercises involving the conducting charged spheres with different radii.
I think that the Italian translators overlooked the fact that the charge distribution is not uniform, in general.
 
  • Like
Likes Delta2
  • #16
261
26
There is one. When close, the potential on each sphere is altered by its proximity to the other sphere. The effect will be greater for the smaller sphere.
According to the approximation in post #6 (which is reckoned to be good both when the spheres are similar in radius and when very different) and taking the case where the ratio of the radii is 1+x, small x, the ratio of the charges approximates 1+3x/2.
Uhm no wait.
I'm thinking of the situation where the spheres touch and then are brought apart, but a "thin extensible wire" maintains them in electrical contact.
This means that the surfaces of the two spheres have always the same potential. This potential probably varies with the distance between the centers of the spheres. Still, I cannot wrap my head around the fact that in the process some charges migrate from one sphere to the other.

From another point of view: start with the two spheres far apart, and connected by the wire. I think that it's safe to say that the charges are uniformly distributed and their ratio equals the ratio of the radii.
Now imagine the wire shortening, and the spheres becoming closer and closer.
My intuition is that the charges on each spherical surface tends to move away from the other sphere, kind of accumulating on the "external" emispheres, and leaving the facing emispheres progressively less charged.
This is all the more true when the spheres touch each other (i.e. when the wire is extremely short, or disappears altogether).
Where does this scenario break, and produce the distribution predicted by Maxwell?

Perhaps the device of the "very thin and long wire" does not really work. But this is kind of an established trick, isn't it?
 
  • Like
Likes etotheipi
  • #17
etotheipi
Gold Member
2019 Award
2,943
1,899
From another point of view: start with the two spheres far apart, and connected by the wire. I think that it's safe to say that the charges are uniformly distributed and their ratio equals the ratio of the radii.
Now imagine the wire shortening, and the spheres becoming closer and closer.
My intuition is that the charges on each spherical surface tends to move away from the other sphere, kind of accumulating on the "external" emispheres, and leaving the facing emispheres progressively less charged.
This is all the more true when the spheres touch each other (i.e. when the wire is extremely short, or disappears altogether).
Yes I also think it is non-intuitive. We "expect" the charges to "repel each other" to the outer sides, but evidently if the formulae presented here are correct there must be a transfer of charge across the wire whilst the separation is being reduced from infinity.

If the distance between the centres of the conducting spheres connected via a conducting wire is changed by a small amount, then we will for a tiny amount of time get a non-zero electric field inside the conductor that will redistribute the surface charges so that the electric field again goes to zero and the volume is again equipotential. But I cannot see anything mathematically stopping charges redistributing through the wire.

So it would appear our intuition is only correct to small order :wink:.
 
  • Like
Likes FranzDiCoccio
  • #18
261
26
Right... the charges repel each other, but on the other hand they "want" to distribute so that the potential inside both spheres is uniform (and the same, if the spheres are electrically connected).
These two effects might be competing.
As you suggest, probably some charges have to redistribute trhough the wire, against the repulsive force, in order to keep the potential uniform inside both spheres.
 
  • #19
etotheipi
Gold Member
2019 Award
2,943
1,899
Right... the charges repel each other, but on the other hand they "want" to distribute so that the potential inside both spheres is uniform (and the same, if the spheres are electrically connected).
These two effects might be competing.
As you suggest, probably some charges have to redistribute trhough the wire, against the repulsive force, in order to keep the potential uniform inside both spheres.
Even though the surface charges might be skewed to the outer hemispheres, during an infinitesimal time until electrostatic equilibrium is again reached the distribution could still cause an electric field to appear inside the wire (pointing in the direction from one sphere to the other), that causes charges to transfer across it.

This isn't so far fetched; in the wire in equilibrium the electric field from the charges on one sphere exactly balance that from the charges on the other sphere. A small change to the configuration would appear to cause a net electric field in the wire for a tiny duration.

Though I am still working through the maths myself, so I do not know if this is a correct interpretation.
 
  • #20
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,861
6,129
Where does this scenario break
Let sphere A be very small.
They start a long way apart, same positive potential. No connecting wire.
Bring them close, but for the moment treat them as insulators, so the charges do not migrate. Each gains in potential from proximity to the other. The potential is uneven on sphere B, but the potential at its centre is the same as that on A.
Now let the charges migrate. Redistribution of the charge on A matters little because it is small, but on B charge moves away from A. This does not change the potential at the centre of B because its charges are still all the same distance from its centre, but it lowers the potential on A.
 
  • #21
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,861
6,129
Even though the surface charges might be skewed to the outer hemispheres, during an infinitesimal time until electrostatic equilibrium is again reached the distribution could still cause an electric field to appear inside the wire (pointing in the direction from one sphere to the other), that causes charges to transfer across it.

This isn't so far fetched; in the wire in equilibrium the electric field from the charges on one sphere exactly balance that from the charges on the other sphere. A small change to the configuration would appear to cause a net electric field in the wire for a tiny duration.

Though I am still working through the maths myself, so I do not know if this is a correct interpretation.
It doesn't sound as though you are using the difference in radii, which must be critical.
 
  • Like
Likes etotheipi
  • #22
etotheipi
Gold Member
2019 Award
2,943
1,899
It doesn't sound as though you are using the difference in radii, which must be critical.
Yes I didn't make specific reference there, I was just trying to say something along the lines that we can still have a transfer of charge between them even if it looks like "intuitively" the charges on each sphere would try to get as far away from each other as possible.

Though I have been messing around with the problem all morning and it has been quite soul-crushing... all I get are integrals that I cannot calculate. I am trying to discard some HOT and get an approximation but now I think I need to have breakfast 😆
 

Related Threads on Charge rearrangement on conducting spheres

Replies
8
Views
11K
Replies
21
Views
16K
Replies
1
Views
2K
Replies
38
Views
10K
Replies
1
Views
2K
Replies
3
Views
802
Replies
2
Views
2K
Replies
8
Views
2K
Replies
4
Views
39K
  • Last Post
Replies
2
Views
1K
Top