Charge distribution on spheres

[Suyash Singh]

Homework Statement

Two conducting spheres having same charge density

and with radius “R” & “2R” are brought in contact and separated by large distance. What are their final surface charge densities ?

Homework Equations

No equation in question.

The Attempt at a Solution

Tried using the fact that distribution of charges on spheres are proportional to their radii but end up with the wrong answer.

Also how to calculate charge distribution for same question but on different shapes.[/B]

 

[gneill]

Hi Suyash Singh,

Welcome to Physics Forums.

According to the forum rules you need to show some details in your attempt at solution. What equations did you use and how did you apply them? What result did you get (even if you believe it to be incorrect)?

 

[haruspex]

Two conducting spheres having same charge density

and with radius “R” & “2R” are brought in contact

The distribution of charge on conducting spheres in direct contact is a very hard problem. It is quite different from two well-separated spheres connected by a wire.
See e.g. http://www.physics.princeton.edu/~mcdonald/examples/twospheres.pdf.

 

[Delta2]

The distribution of charge on conducting spheres in direct contact is a very hard problem. It is quite different from two well-separated spheres connected by a wire.
See e.g. http://www.physics.princeton.edu/~mcdonald/examples/twospheres.pdf.

Where the following reasoning goes wrong?
"During contact the spheres exchange electric charge so that the final equilibrium is when both have the same potential. The potential of each sphere is given by $$V=K\frac{Q}{R}$$" so the equilibrium is when $$\frac{Q}{R}=\frac{Q'}{R'}$$"

 

[haruspex]

The potential of each sphere is given by

That's for a sphere in isolation. Each sphere's potential is affected by the charge on the other sphere, and the distributions will not be uniform.

 

[Delta2]

That's for a sphere in isolation. Each sphere's potential is affected by the charge on the other sphere, and the distributions will not be uniform.

Ok I see but , as also the paper that you linked says, this reasoning is correct for the zero order approximation. This problem probably is given in high school, so for high school physics this approximation can be taken as correct?

 

[haruspex]

Ok I see but , as also the paper that you linked says, this reasoning is correct for the zero order approximation. This problem probably is given in high school, so for high school physics this approximation can be taken as correct?

According to Suyash, that got the wrong answer.

 

[Suyash Singh]

well the answer is 5/3sigma and 5/6 sigma

Where the following reasoning goes wrong?
"During contact the spheres exchange electric charge so that the final equilibrium is when both have the same potential. The potential of each sphere is given by $$V=K\frac{Q}{R}$$" so the equilibrium is when $$\frac{Q}{R}=\frac{Q'}{R'}$$"

i did this in my exam too but it doesn't give the correct answer.

 

[Suyash Singh]

by the way its school physics so only basic formulas would be required and everything would be ideal.
I also did this through "charging by contact method" but that also doesn't seem to work

 

[haruspex]

it doesn't give the correct answer.

I believe it does give the answer you quote. Please post your working.

 

[Suyash Singh]

here's what i did
let $ be sigma(surface charge density)
k(q)/r=k(q')/2r
whch can be written as
$ pi r r/r=$ pi 4 r r/2r
-$=0
which doesn't make sense

 

[haruspex]

here's what i did
let $ be sigma(surface charge density)
k(q)/r=k(q')/2r
whch can be written as
$ pi r r/r=$ pi 4 r r/2r
-$=0
which doesn't make sense

They have the same surface charge density and different potentials before they are brought into contact.
When they come into contact, both densities change and the potentials become equal. Only then can you apply k(q)/r=k(q')/2r.

 

[rude man]

The distribution of charge on conducting spheres in direct contact is a very hard problem. It is quite different from two well-separated spheres connected by a wire.
See e.g. http://www.physics.princeton.edu/~mcdonald/examples/twospheres.pdf.

Irrelevant since the spheres are remote from each other after contact.

 

[rude man]

Where the following reasoning goes wrong?
"During contact the spheres exchange electric charge so that the final equilibrium is when both have the same potential. The potential of each sphere is given by $$V=K\frac{Q}{R}$$" so the equilibrium is when $$\frac{Q}{R}=\frac{Q'}{R'}$$"

Your approach is OK as far as it goes but you also need to relate charges on the two spheres before and after contact. There is Q and Q' before contact and there is different Q and Q' after contact, but the sum of before & after is constant.

 

[haruspex]

Irrelevant since the spheres are remote from each other after contact.

The charges become redistributed while in contact. Once they lose contact, those charges are fixed. Only the way each charge is distributed changes after that.
To make the problem work, it should have specified that they are kept well apart but connected by a thin wire . That is a very standard formulation.

Your approach is OK as far as it goes but you also need to relate charges on the two spheres before and after contact. There is Q and Q' before contact and there is different Q and Q' after contact, but the sum of before & after is constant.

... as I wrote, and Suyash acknowledged, in post #12.

 

[rude man]

The charges become redistributed while in contact. Once they lose contact, those charges are fixed. Only the way each charge is distributed changes after that.
To make the problem work, it should have specified that they are kept well apart but connected by a thin wire . That is a very standard formulation.

The wire is not needed; why should the potentials change relatively once they have made contact?
EDITed again.
@haruspex I will send you a private note, see what you think.

 

[haruspex]

The wire is not needed; why should the potentials change relatively once they have made contact?

While in contact, they share the charge, distributed so as to have the same potential on both surfaces. But what is that potential, and how much charge is on each sphere? This is a tough problem, and as I understand it there is no exact solution.
Did you follow the link I posted?

 

[rude man]

... as I wrote, and Suyash acknowledged, in post #12.

??

 

[rude man]

I wind up with the assumption that the potential, upon contact, whatever it may be, is unchanged for both spheres after they're separated. I still can't rigorously defend that assumption but I'm pretty certain it's assumed in the problem statement, especially since it's supposed to be below college engineering level physics. Connecting the two spheres by a wire as haruspex describes would remove the need for that assumption.

 

[haruspex]

I wind up with the assumption that the potential, upon contact, whatever it may be, is unchanged for both spheres after they're separated.

Very unlikely. While in contact, the potential at each is the result of the charge distribution over both spheres.
Have you read the paper I linked to yet?
Take a look at eqn 4.1 at http://rspa.royalsocietypublishing.org/content/royprsa/468/2145/2829.full.pdf.

Edit: using that equation I get that for spheres radius R, 2R and total charge Q the potential while in contact is $\frac Q{2R\ln 3}$. The individual charges are $\frac Q2\left(1-\frac{\pi\sqrt 3}{9\ln 3}\right)$ and $\frac Q2\left(1+\frac{\pi\sqrt 3}{9\ln 3}\right)$

 

[rude man]

Very unlikely. While in contact, the potential at each is the result of the charge distribution over both spheres.
Have you read the paper I linked to yet?
Take a look at eqn 4.1 at http://rspa.royalsocietypublishing.org/content/royprsa/468/2145/2829.full.pdf.

Edit: using that equation I get that for spheres radius R, 2R and total charge Q the potential while in contact is $\frac Q{2R\ln 3}$. The individual charges are $\frac Q2\left(1-\frac{\pi\sqrt 3}{9\ln 3}\right)$ and $\frac Q2\left(1+\frac{\pi\sqrt 3}{9\ln 3}\right)$

Thanks, yes, I looked at it now & I have to compute a couple of d/dx [gamma(x)] at x = 1/3 and x=2/3
but it does look like [γ + ψ(1/3)] / [γ + ψ(2/3)] ≠ 2 is likely.
which it would have to if the potentials did not change. I'll post again when I do the gamma function derivatives.
Thanks very much for this paper!

 

[haruspex]

it does look like [γ + ψ(1/3)] / [γ + ψ(2/3)] ≠ 2

Yes, the charge ratio is about 3.45.

 

[rude man]

Yes, the charge ratio is about 3.45.

Thanks for saving me the computation.
Interesting is that the answer to the stated problem is strictly a functon of total initial charge and ratio of radii. No potential computations needed anywhere.

 

[haruspex]

Thanks for saving me the computation.
Interesting is that the answer to the stated problem is strictly a functon of total initial charge and ratio of radii. No potential computations needed anywhere.

Not sure what you mean. Maxwell would have had to work in terms of potentials to arrive at eqn 4.1.

Did you look at how the equations are obtained in that paper? It uses the method of images to show that (as far as the field outside the spheres is concerned) the charges on the spheres can be represented by infinite sequences of point charges along the axis. I assume Maxwell did the same,

 

[rude man]

Not sure what you mean. Maxwell would have had to work in terms of potentials to arrive at eqn 4.1.

Yes, but we didn't! All we needed was eq. 4.1 divided by eq.4.3 which he did for us, then I worked solely with Qa/Qb before & after contact. The potential canceled for that ratio.

I trusted Maxwell so did not look further into that derivation. In fact I don't remember seeing it in the paper, it appeared to be a given.

 

[haruspex]

I trusted Maxwell so did not look further into that derivation. In fact I don't remember seeing it in the paper, it appeared to be a given.

I did not mean specifically the derivation of 4.1. Both papers I linked to use the method of images, and 4.1 was very likely obtained the same way.

 

[rude man]

I did not mean specifically the derivation of 4.1. Both papers I linked to use the method of images, and 4.1 was very likely obtained the same way.

The only part of the paper I found relevant was eqs. 4.1 and 4.3, which were derived by maxwell aroud 1891. I was not interested in the rest of the paper which seemingly dealt with attractive & repulsive forces. Thanks again for finding this paper.