What is the potential between 2 concentric spheres?

[Red]

Homework Statement

Consider two concentric spherical conducting shell. The inner sphere has radius r₁, potential V₁, while the outer sphere has radius r₂, potential V₂. Find the potential at the center of these two sphere, at r₀=(r₁+r₂)/2.

2. The attempt at a solution
I tried to use method of images, by modelling there to be a charge q₁ at the center of the inner sphere and q₂ at the center of the outer sphere. Since both sphere are concentric, q₁ and q₂ are at the same location. So I attempt to add up q₁ and q₂ (which I now know is wrong), and then find the potential at r₀. My answer is double the correct answer.

 

[STEMucator]

You can treat each concentric sphere like a point charge and calculate the potential of each one using $V = k \frac{q}{r}$.

You have to add the potentials at the point to get the net potential. Could you show your working for $V_1$ and $V_2$?

 

[Red]

Sure. I let V₁=-kq₁/r₁, V₂=-kq₂/r₂. I made both q the subject and sum them up, q₀=q₁+q₂=(-r₁V₁-r₂V₂)/k. Then I proceed to find the potential at center of both sphere, V₀=-kq₀/r₀=(r₁V₁+r₂V₂)/r₀.

 

[STEMucator]

So both charges are negative?

How far away is the inner point charge from the point?

How far away is the outer point charge? Those are the radial distances you should be concerned with.

Then sum $V_{net} = V_1 + V_2$.

 

[haruspex]

You can treat each concentric sphere like a point charge and calculate the potential of each one using $V = k \frac{q}{r}$.

Not so fast - you cannot treat a uniform shell of charge as a point charge for calculating potentials inside the shell.

I let V1=-kq1/r1, V2=-kq2/r2.

The potential at the outer shell is not solely due to the charge on the outer shell. There is a contribution from the inner shell.

 

[Red]

Yes exactly, the potential at the outer shell is related to the charge that contribute to the inner shell. Haruspex how can I find the mean potential at the center of the two shells?

 

[haruspex]

Yes exactly, the potential at the outer shell is related to the charge that contribute to the inner shell. Haruspex how can I find the mean potential at the center of the two shells?

There may be a quicker way, but this way looks safe:

Suppose the charges are q1, q2. Compute the potentials that should result. What equations does that give you?

 

[Red]

I don't quite get you. If I sum up q1 and q2 then I get V0=(r1V1+r2V2)/r0.

 

[ehild]

Assume that the charge on the inner shell is q1 and on the outer shell is q2. Applying Gauss' Law, you can determine the electric field E(r) between the shells (E1) and outside the bigger shell (E2).
Remember that the electric field is gradient of the potential: you get the potential difference between two points if you integrate -E(r) between those points. Choose V(∞)=0 and integrate E2 between r2 and infinity. Integrate also E1 between r1 and r2. You get two equations which relate the unknown charges to the potentials V1 and V2. If you know the charges it is easy to get the potential at r=(r1+r2)/2

ehild

 

[Red]

Hi ehild, thank you for your response, however the charge on the shells are not known. What is known is that the outer shell has potential V2 and the inner shell has potential V1. How do I find the potential at the center between these two shells?

 

[haruspex]

I don't quite get you. If I sum up q1 and q2 then I get V0=(r1V1+r2V2)/r0.

No, I mean compute the potentials that ought to result at the two shells. Equating those to the given potentials allows you to deduce the charges.
Note that the two potentials are not simply q1/r1 and q2/r2, which you seem to be assuming.

 

[ehild]

Hi ehild, thank you for your response, however the charge on the shells are not known. What is known is that the outer shell has potential V2 and the inner shell has potential V1. How do I find the potential at the center between these two shells?

Find the relation between the charges and the potentials. From those, you can determine the charges.

Supposing the inner shell has q1 charge. What is the electric field around it in terms of q1 if r1<r<r2? Apply Gauss' Law.

ehild

 

[Red]

Thank for your responses. I think there is an misunderstanding. Maybe I should ask the question in a slightly different way: A potential V_H is applied between 2 spherical conducting shell. The inner sphere has radius r₁, while the outer sphere has radius r₂. What is the potential at the center of these two sphere, at r₀=(r₁+r₂)/2?

 

[Orodruin]

Assume that the shells have charges $q_1$ and $q_2$, respectively. What would be the potential between the shells in this situation? (Noting that you are inside one of the shells.)
If you have an expression for the potential given the two charges, then you can apply the boundary conditions to this, i.e., the potential at the outer shell has to be $V_2$ and the inner $V_1$. This gives you two equations and two unknowns, so you will be able to deduce the charges and thus the potential. After this you should be able to simply plug in whatever $r$ you want into this.

 

[haruspex]

Thank for your responses. I think there is an misunderstanding. Maybe I should ask the question in a slightly different way: A potential V_H is applied between 2 spherical conducting shell.

It doesn't matter how the shells have the given potentials. It might as well be via an applied charge, so you can solve it the way ehild, Orodruin and I are suggesting. We're all telling you the same method, so how about trying it?

 

[ehild]

Thank for your responses. I think there is an misunderstanding. Maybe I should ask the question in a slightly different way: A potential V_H is applied between 2 spherical conducting shell. The inner sphere has radius r₁, while the outer sphere has radius r₂. What is the potential at the center of these two sphere, at r₀=(r₁+r₂)/2?

You misunderstand the problem. The two shells are concentric, the radii are given, r1 and r2. The potential is given for both spheres; V1 and V2 . The problem asks the potential at distance ro=(r1+r2)/2 from the common centre of the spheres. In the centre, the potential is the same as on the inner sphere, that is V1.

If you do not want to use charges, remember that the electric potential U obeys the Laplace equation at points where the charge density is zero. Because of the spherical symmetry of the problem, U depends only on the radius r, and $$\frac{1}{r^2}\frac {\partial }{\partial r}\left(r^2 \frac {\partial U}{\partial r}\right)=0$$ with the boundary conditions U(r1)=V1, U(r2)=V2.
See http://en.wikipedia.org/wiki/Laplace_operator
What is the solution for U(r) between the shells r1<r<r2?

ehild

 

[Orodruin]

It doesn't matter how the shells have the given potentials.

Or, put in another way, as long as you are not putting charges between the shells, any charge configuration that fulfills the boundary conditions is going to give you the same potential. You might as well select a surface charge on the shells. Or you could assume a point charge in the common center of the spheres and a surface charge on a shell with radius $R \gg r_2$. Both approaches will give you the same form for the potential between the shells, which can then be adjusted to accommodate the boundary conditions.

The alternative is to do what ehild suggests and bite into the Laplace equation with spherical symmetry in spherical coordinates. This requires at least some knowledge about differential equations and how to solve them (in particular differential equations of Euler type).

Note to self: You have been spending too much time on PF when you start writing ## instead of $ in your LaTeX source ...

 

[D.R.U]

Potential in inner part of spear is always zero due to equal distribution of charge. ..Right?...but outside of spear there is net potential ...okey so.

Answer is...Potential in between spear is only due to inner spar...outer spear effect is zero...But outside of whole spear there is net potential occur... So

Answer:
(1).inside the inner spear,potential is zero.
(2).between two spear, potential is only due to inner spear.
(3).outside the whole spear there is potential present due to both spears.

 

[ehild]

Potential in inner part of spear is always zero due to equal distribution of charge. ..Right?...but outside of spear there is net potential ...okey so.

The potential inside a conductor is constant.
And what do you mean on spear? The problem is about concentric spherical shells.

ehild

 

[Orodruin]

The potential inside a conductor is constant.
And what do you mean on spear? The problem is about concentric spherical shells.

If I am going to attempt a translation of DRU's post, I do believe "spear = sphere" and "potential = field" ...

 

[Orodruin]

Homework Statement

A potential V_H is applied between 2 spherical conducting shell. The inner sphere has radius r₁, while the outer sphere has radius r₂. Find the potential at the center of these two sphere, at r₀=(r₁+r₂)/2.

2. The attempt at a solution
I tried to use method of images, by modelling there to be a charge q₁ at the center of the inner sphere and q₂ at the center of the outer sphere. Since both sphere are concentric, q₁ and q₂ are at the same location. So I attempt to add up q₁ and q₂ (which I now know is wrong), and then find the potential V₀ at r₀. I get (r₁V₁+r₂V₂)/r₀. My answer is double the correct answer.

Since you reposted this in the Advanced forum, let me bring it back here.

Your main problem seems to be that you are assuming that the potential of both charged shells behave as $q/r$. This is only true as long as you are outside of the shell and the only shell you are outside of is the smaller one. Inside a spherical shell, the potential is constant (see http://en.wikipedia.org/wiki/Shell_theorem - it is described for gravity but applies just as well to electrostatics) and proportional to $q/R$, where $R$ is the (constant) shell radius and not the coordinate $r$

What does this tell you about the potential between the two shells?

 

[Red]

Thank for all of your responses, I really appreciate them. I am quite confused now. I let the inner shell has a surface charge density of ρ and using Gauss Law I found the electric field between the shells. The electric field is E(r)=r₁²ρ/ε₀r². How should I continue to find the potential at r0? I attempted to take an integration from r1 to r0 but I cannot get the answer.

 

[Orodruin]

Since you are not showing exactly what you did I am going to take a stab in the dark here. Did you do the following?

1. Compute the $\rho$ needed to give a single shell potential $V_1$.
2. Use that to get the field.
3. Attempted to use that field to get to the potential you want.

The above procedure fails because your setup is not a single shell. The potential at r1 is also going to have a contribution from the outer shell so in order to get the correct potential of the inner shell you need to account for this.

However, it is true that the field between the shells is going to be given by the inner charge only. Let this charge be an unknown and integrate to get the potential at an arbitrary r between the shells. What is the result? (Do not assume the potential goes to zero at infinity, infinity is not part of your domain.)

 

[Orodruin]

And yet another method:

You know what the potential of a point charge in the center is. What is the resulting potential difference between r1 and r2? Can you think of any way of changing the potential without changing the fields? (And thus without changing the potential difference between r1 and r2.)

 

[D.R.U]

The potential inside a conductor is constant.
And what do you mean on spear? The problem is about concentric spherical shells.

ehild[/QUOT
sorry for English...there is potential difference and sphere...

 

[D.R.U]

Potential at center of both sphere is (v1 + v2)...

 

[D.R.U]

Potential at center of both sphere is (v1 + v2)...
Where v1= potential incide outer sphere and v2 for inner sphere.

 

[Orodruin]

Potential at center of both sphere is (v1 + v2)...
Where v1= potential incide outer sphere and v2 for inner sphere.

This is not the answer to the intended problem. He wants to compute the potential at $r = (r_1+r_2)/2$, i.e., halfway between the shells.

Furthermore, the potential inside of the inner shell is just $V_1$, assuming there inner shell at $r=r_1$ is kept at this potential and there are no charges inside. If the potential was $V_1+V_2$, the potential would not be continuous at $r=r_1$ ...

Also, this is the homework forum, we are here to try to help people solve problems, not solve the problems for them.

 

[ehild]

Thank for all of your responses, I really appreciate them. I am quite confused now. I let the inner shell has a surface charge density of ρ and using Gauss Law I found the electric field between the shells. The electric field is E(r)=r₁²ρ/ε₀r². How should I continue to find the potential at r0? I attempted to take an integration from r1 to r0 but I cannot get the answer.

Take the integral of E(r) from r1 to a general r, to get the potential function U(r).

As $-\nabla U = E(r)$

$$-\int_{r_1}^{r} \nabla U(r') dr' =V_1-V(r) =\int_{r_1}^{r}E(r')dr'$$

and use the condition that $$V_1-V_2 =\int_{r_1}^{r_2}E(r')dr'$$

to get the unknown ρ.

 

[haruspex]

Thank for all of your responses, I really appreciate them. I am quite confused now. I let the inner shell has a surface charge density of ρ and using Gauss Law I found the electric field between the shells. The electric field is E(r)=r₁²ρ/ε₀r². How should I continue to find the potential at r0? I attempted to take an integration from r1 to r0 but I cannot get the answer.

Introducing charge densities is an unnecessary complication. It isn't a hard problem once you get the concepts straight.

There are standard formulae for the potential at a given distance from a uniformly charged spherical shell, which I assume you know. It falls into two cases, according to whether the point is inside or outside the shell.

Let the charges on the spheres be q1, q2. You have four resulting potentials to calculate:
- the potential at r1 resulting from q1
- the potential at r2 resulting from q1
- the potential at r1 resulting from q2
- the potential at r2 resulting from q2

From the above potentials, find
- the total potential at r1 (which equals V1)
- the total potential at r2 (which equals V2)

That gives you two equations with two unknowns. From these you can express q1, q2 as functions of V1, V2.

 

[Orodruin]

It never ceases to amaze me how you can say a relatively basic thing in so many ways.

If you have not yet solved it, since you seem to be at least somewhat familiar with integration and therefore should be with differentiation, I would suggest going back to ehild's post #16. You may or may not prefer the differential equation on the form
$$
r^2 U''(r) + 2r U'(r) = 0,
$$
with boundary conditions $U(r_1)=V_1$, $U(r_2)=V_2$. Once done it will be easier to discuss the solution and the interpretation of it.

 

[STEMucator]

It never ceases to amaze me how you can say a relatively basic thing in so many ways.

If you have not yet solved it, since you seem to be at least somewhat familiar with integration and therefore should be with differentiation, I would suggest going back to ehild's post #16. You may or may not prefer the differential equation on the form
$$
r^2 U''(r) + 2r U'(r) = 0,
$$
with boundary conditions $U(r_1)=V_1$, $U(r_2)=V_2$. Once done it will be easier to discuss the solution and the interpretation of it.

The S.I. unit system and the concepts of mathematics are a privilege to learn, and to apply.

That differential equation solution is quite interesting, but the simplicity of haruspex's post would probably be the elementary way of visualizing this.

 

[Orodruin]

The SI unit system has nothing to do with the matter. The problem as posed does not require a system of units to solve. Are you perhaps rather referring to the use of basic solutions such as that of a point charge or a spherical shell to solve the problem? My point is that several people, including ehild, haruspex, and myself, have been trying to find a way of getting that message through for the better part of two full pages of posts now. Solving the differential equation and use the solution to discuss the physics of it afterwards is my if-all-else-fails approach.

 

[ehild]

It never ceases to amaze me how you can say a relatively basic thing in so many ways.

If you have not yet solved it, since you seem to be at least somewhat familiar with integration and therefore should be with differentiation, I would suggest going back to ehild's post #16. You may or may not prefer the differential equation on the form
$$
r^2 U''(r) + 2r U'(r) = 0,
$$
with boundary conditions $U(r_1)=V_1$, $U(r_2)=V_2$. Once done it will be easier to discuss the solution and the interpretation of it.

The equation in Post #16 is easier to solve for r1<r<r2:

$\frac{1}{r^2}\frac {\partial }{\partial r}\left(r^2 \frac {\partial U}{\partial r}\right)=0$

Multiply by r²:

$\frac {\partial }{\partial r}\left(r^2 \frac {\partial U}{\partial r}\right)=0$

Integrate once, the right side is a constant: $r^2 \frac {\partial U}{\partial r}=A\rightarrow \frac {\partial U}{\partial r}=\frac{A}{r^2}$

Integrate again :$U=-\frac{A}{r}+B$.

Use the boundary conditions:

$V_1=-\frac{A}{r_1}+B$
$V_2=-\frac{A}{r_2}+B$

It is two equations for the constant A and B. Solve, substitute into the equation for U(r): $U=-\frac{A}{r}+B$. You can calculate the potential at any point between the shells.

ehild

 

[Orodruin]

The equation in Post #16 is easier to solve for r1<r<r2:

I guess this depends on your definition of "easy" and pre-knowledge. Recognizing it as an ODE of Euler type, it is just about insering $r^k$ and solving the resulting $k(k+1)=0$.

That said, neither method is really hard.