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# Help me out with finding electric potential in this task

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#### alitavakol

##### Member level 2
a point charge q is placed a distance d from the center of a conducting spherical shell of radius r (r < d). total charge of the sphere is zero. find the electric potential everywhere. justify your answer.

Re: method of images

According to your statement, the point charge is placed outside of the sphere. Besides, "total charge of the sphere is zero", and, therefore, the sphere does NOT contribute anything. The potential is completely decided by that point charge, which is q/(4*pi*epsilon0*r) and where r is the distance from that point charge to any point.

Here is my question for you. Why does that "total charge of the sphere is zero" jump into the picture? My guess is that you really want to put another negative point charge inside the sphere so that the potentials over the sphere generated by the two charges cancel each other.

Re: method of images

Hi

The sphere does influence on field. Hope that this link can help.
**broken link removed**

Regards

Re: method of images

Be careful. we don't know the potential of the sphere. all we know is that its total charge is zero. also, when we put a point charge near it, the charge distribution on the sphere is not then uniform.

Re: method of images

zajbanlik
Can you tell me why the sphere has influence on the potential when there is only one point charege in the space? The link you refered all shows that at least two point charges are in the picture. Find another link?

alitavakol,
Can you explain what does this mean, "we don't know the potential of the sphere. all we know is that its potential is zero"? You don't know its potential but you know its potential is zero?

Re: method of images

steve10 said:
zajbanlik
Can you tell me why the sphere has influence on the potential when there is only one point charege in the space? The link you refered all shows that at least two point charges are in the picture. Find another link?

alitavakol,
Can you explain what does this mean, "we don't know the potential of the sphere. all we know is that its potential is zero"? You don't know its potential but you know its potential is zero?
Hi

Well the sphere has some positive charge on one side and some negative charge on some other side (its not uniform). Like alitavakol has written try to find potential, and read carffuly document from previous link. You will see the nature of that second charge (this second charge is way of modeling presence of sphere). Or do the google search with keywords "method of images".

Regards

Re: method of images

steve10 said:
zajbanlik
alitavakol,
Can you explain what does this mean, "we don't know the potential of the sphere. all we know is that its potential is zero"? You don't know its potential but you know its potential is zero?
sorry! I have fixed it.

Re: method of images

I am sorry that I am almost completely confused. Let's see if we can solve this problem.

zajbanlik,
When you say "Well the sphere has some positive charge on one side and some negative charge on some other side ...", what information (from alitavakol) are you based on? I only see source from his very first post "a point charge q is placed a distance d from the center of a conducting spherical shell of radius r (r < d). total charge of the sphere is zero ..." which says nothing about the charge one side or other of the sphere. After he made change to his post, he only says " ...all we know is that its total charge is zero..", my understanding is that the charge is spread over the sphere, but not inside or outside ... Are we talking about alitavakol's problem or the problems in your link?

alitavakol,
Can you expain what this means, "when we put a point charge near it, the charge distribution on the sphere is not then uniform"? If you put the charge only "near" the sphere, the charge on the sphere is still zero at any point. Why does this sphere have anything to do with the potential?

After you made change to your post, I see you get infinitely many solutions. Here is why. you put a positive charge q0 and a negative charge -q0 on the sphere, which will make the total charge zero. But notice that you can put them anywhere on the sphere. Therefore, the total potential generated at any point will be the sum of the three potentials (the other is the one with charge q you put at somewhere which has a distance d from the center of the sphere). Since you have infinitely many choices of the location of the two charges on the sphere, you get infinitely many solutions which look like (q/r+q0/r1-q0/r2)/(4*pi*epsilon0) and where r, r1 and r2 are the distances of the three charges from any point.

Re: method of images

Hello

OK Let's be more detail.
The charge on sphere is not uniform because of the influence of charge q which is outside the sphere (it is in total 0, but that doesn't means that schould be 0 in every point). Don't forget one thing, we are having conductive sphere. that means electrical fild on spere surcace qan only be radial ( in all other axes field is zero). Thats mean we have to have some charge redistribution on sphere to satisfy that.
That's why when you put conductive sphere in some field it changes the distribution of the field.
If you want to use some mathematical equations to calculate field without using images method it would be difficult.
Hope that now is more clear what I wanted to say

Regards

Re: method of images

zajbanlik said:
Hello

OK Let's be more detail.
The charge on sphere is not uniform because of the influence of charge q which is outside the sphere (it is in total 0, but that doesn't means that schould be 0 in every point). .....
Regards

Obviously, we are going back and forth between those terms, charge, potential, electric field. This is what he said, "Be careful. we don't know the potential of the sphere. all we know is that its total charge is zero ..." Does it mean we have charge on the sphere? The potential and charge are different concepts. The influence is about the potential,but not charge.

I am pretty sure he is trying to find the Green's function for the sphere. However, the Green's function depends on the boundary conditions. The only thing that keeps confusing me is that it's very hard to figure out what boundary condition he is talking about, First (Dirichlet) or second (Neumann)?

Re: method of images

Well

Let's for a moment forget about this document where is exolained how to calculate the field, potential etc. and lets concentrate on some basic stuff to see where missunderstanding is.
1. Do You agree that total charge of the ball doesnt change because of the presence of external charge?
2. Do You agree that in presence of external (external concerning sphere) charge q, charge of the sphere is not uniform? I mean in total is 0, but in some points is not 0.
3. Do You agree with me that this ununiform charge change the field (especialy around sphere)?
4. Do You agree with me that this change of the field change potential ?
5. Do You agree with me that potencial on entire sphere is the same (I mean he doesnt change)?

So lets see where is a point where we don't agree so that we can find solution.

Regards

Re: method of images

Here are my responses to your questions:

1. yes, I do.
2. yes, I agree if there is charge on the sphere.
3. Don't quite understand your question. Charge will produce an electric field. If you got new charges somewhere, the field will change.
4. Yes, adding any new charges will change the potential.
5. This sounds contradicting to #4. That one says the potential is going to change while you adding new charge while this one says it does not change, which I don't agree with.

Well, if you bring a whole electromagnetic book here and make up a list, I can still check your items one by one, but those don't serve to solve the problem. I would always want to stick to the information that the first poster provides. Our trouble is that he stops providing further information and we don't even have an idea if there is charge on the sphere. This looks just like that a few guys are enthusiastically chasing a pretty girl when the girl suddenly dies for not having enough blood to make her live in this world. The poor men completely lose the direction and start to blame each other... Let's have a break, buddy, and perhaps you like to play some softball...

Re: method of images

Hm
May be you are right lets forget it. By the way I prefere basketball

P.S. If you find something interesting about this subject post it

Re: method of images

I accept all terms, which steve10 said.
I want to solve my problem. if instead of saying that the total charge of the sphere is zero, we say that the potential on the sphere is V0, we can put two image charges in the sphere, satisfying the current boundary conditions. one for vanishing the potential on the sphere, and one (at the center) for rendering the potential on the sphere V0.
but here, we can not find the latter charge. because we don't know V0. it is accepted that q changes the potential on the sphere*. but I can not find V0.
Can V0 be determined?

--------------------------------------------
* reason for zajbanlik: if we connect the sphere to the ground, and then q slowly approaches the sphere, charge on the sphere will flow through the wire to the ground. so, if we disconnect the wire, some force from q will impose the sphere, changing its potential.

Added after 2 hours 40 minutes:

I solved the prolem!
By use of superposition, I divide the configuration in two parts:
1- the point charge q and a grounded conducting spherical shell.
2- a spherical shell with uniform charge distribution q (not grounded).

I claim that the superposition of these two configurations equivalents the problem. because if you solve part 1 by method of images, you will find that the total charge flowed onto the sphere from ground is -q. potential everywhere is also can be determined. every electromagnetics book has solved this problem.
if these two parts coincide, you see that the total charge of the sphere is zero and so the potential is determined everywhere.
if you (steve10 and zajbanlik) agree, I will say helped me.

Re: method of images

COngratulations!

That looks like a lot of work and will certainly make us all feel better.

Like everyone else along this thread, I tried to help, but I don't think I did. Please do NOT bother to say that.

### alitavakol

Points: 2
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Re: method of images

Well I learned one very important thing. Don't try to answer on such questions without JPG image which ilustrate the problem . You schould post the image with some extra explanation and the we could find solution very easy.

Regards

### alitavakol

Points: 2
Helpful Answer Positive Rating
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