# Potential for a system of a solid sphere and spherical shell

## Homework Statement

A metal sphere with radius a is supported on an insulating stand at the center of a hollow, metal spherical shell with radius b. There is charge +Q on the inner sphere and charge -Q on the outer shell. Take the potential V to be zero at infinite separation.

Calculate the potential V(r) for:

i) r < a
ii) a < r < b
iii) r > b

## Homework Equations

1) $$E = -\frac{\partial V}{\partial r}$$

2) $$V = \int E d\ell$$ (along radial direction)

## The Attempt at a Solution

(iii) I already know; V is obviously 0 outside both shells (since E is also zero outside both, as a Gaussian sphere drawn beyond r(b) will enclose Q-Q = 0 charge).

I derived (ii) as a problem in lecture last week.

This leaves only (i) as the problem. The attempt was to use the second of the two equations up top, first finding the electric field so it can be applied:

Applying Gauss' law, we can find that the Electric field inside the inner sphere is:

$$\frac{q'}{q} = \frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi a^{3}}$$

$$q' = q\frac{r^{3}}{a^{3}}$$

From there, using a Gaussian sphere with a radius r < a:

$$E(4\pi r^{2}) = \frac{q\frac{r^{3}}{a^{3}}}{\epsilon_{0}}$$

$$E = \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}}$$

And then applying the equation for potential difference:

$$\int E d\ell = -\Delta V$$

Where dl = dr, choosing a radial path from infinity to r.

$$\int \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}} dr = -\Delta V$$

$$\frac{1}{4\pi \epsilon_{0}} \frac{qr^{2}}{2a^{3}} = \Delta V = V_{r} - V_{\infty} = V_{r}$$

I'm suspicious of this solution though, as it doesn't reduce to
$$\frac{1}{4\pi \epsilon_{0}} \frac{q}{a}$$
For the case where r = a.

Did the mistake lie in how I calculated the electric field, the technique I used to determine potential, or something else altogether?

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BvU
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The sphere is made of metal. Metals conduct, so if there were a potential difference, charge would start moving until there is no more potential difference. Charges of equal sign repel each other, so they sit on the very outside of the inner shell, nicely equally distributed.

The sphere is made of metal. Metals conduct, so if there were a potential difference, charge would start moving until there is no more potential difference. Charges of equal sign repel each other, so they sit on the very outside of the inner shell, nicely equally distributed.
That the inner sphere is on an insulating stand doesn't make a difference here?

BvU
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Insulators can be considered ideal in this kind of exercises, I would say. So there is no influence.

Insulators can be considered ideal in this kind of exercises, I would say. So there is no influence.
I don't understand. Do you mean that because the insulator is ideal, it would have no effect at all on the sphere it supports?

BvU
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That is the idea. You sense that I am a little careful in formulating. An insulator does not allow charge to flow. But an insulator can have a degree of polarizability that influences the field. Since you are not provided with any information on the insulator at all, you may assume its effect is negligible (for example because it is relatively thin)

Things change when e.g. the entire space between sphere and shell is filled with a polarizable dielectric material; then you have a capacitor with a higher capacity, because the polarization of the material counteracts the electric field (thus lowering the potential difference).

This leaves only (i) as the problem. The attempt was to use the second of the two equations up top, first finding the electric field so it can be applied:

Applying Gauss' law, we can find that the Electric field inside the inner sphere is:

$$\frac{q'}{q} = \frac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi a^{3}}$$

$$q' = q\frac{r^{3}}{a^{3}}$$

From there, using a Gaussian sphere with a radius r < a:

$$E(4\pi r^{2}) = \frac{q\frac{r^{3}}{a^{3}}}{\epsilon_{0}}$$

$$E = \frac{1}{4\pi \epsilon_{0}} \frac{qr}{a^{3}}$$
The inner metal sphere is a conductor ,hence electric field inside the sphere(r<a) is zero .

The electric field is present only between the outer shell and the inner metal sphere .

Another approach and an easier one is to find the potential within the sphere using superposition of potentials due to the charge on the inner metallic sphere and the outer metallic shell .

BvU
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2019 Award
@Tanya: if you were my student, I would compliment you. However, if you were my teacher I would not. CW doesn't benefit from answers just dumped in his/her lap. Fortunately, ii) and iii) were already dealt with, and we were well under way with i). I was quite happy with CW's curiosity after the insulator issue. Admittedly a sidetrack. And at the end we would also have adressed CW's unhappiness (suspicion) with the discrepancy found at r=a

@Tanya: if you were my student, I would compliment you. However, if you were my teacher I would not. CW doesn't benefit from answers just dumped in his/her lap. Fortunately, ii) and iii) were already dealt with, and we were well under way with i). I was quite happy with CW's curiosity after the insulator issue. Admittedly a sidetrack. And at the end we would also have adressed CW's unhappiness (suspicion) with the discrepancy found at r=a
I checked earlier today with the professor, who explained the insulator discrepancy, and it was evidently just the text's way of creating the "solid sphere within a spherical shell" scenario without actually having the sphere float in the center unsupported, with the stand having no impact on the system (which, upon re-reading, appears to be what you were saying). I just didn't catch on to this, as based on the unrealistic idealization of most problems in this textbook, having a sphere suspended within a shell unsupported didn't seem like much of a stretch.

But based on this, the answer I determined was:

$$\Delta V = V_{r} - V_{\infty} = V_{r}$$

$$V_{r} = \frac{1}{4\pi \epsilon_{0}} \int \frac{dq}{r}$$

And since r < a, and there is no charge within the inner sphere, dq = 0, and hence, V = 0 for r < a.

To confirm, is this sound reasoning?

@BvU: I am not sure what prompted your response in post#8 .Please understand this very clearly that I only chipped in when i found that OP wasn't told explicitly where and how he/she was committing mistake in post#1 .I simply pointed out that mistake.That is how the forum works.I do not have this habit of poking my nose just for the sake of proving some point.

@Tanya: if you were my student, I would compliment you. However, if you were my teacher I would not.
Well...I am glad ,I am neither of the two.

CW doesn't benefit from answers just dumped in his/her lap.
Did I dump the answers ? Surely not .I discussed about a key concept (that electric field inside a metal conductor is zero) which you should have pointed straightaway .Instead you were discussing about the insulator which hardly has any role to play in the problem except that it prevents a conducting path from the sphere to earth.

Second thing I discussed was the principle of superposition of potentials .There are times when you need to give further insight into a problem.That was an alternative approach.The mathematics was supposed to be done by OP.

In any case,I didn't do any work for OP. If you feel I have done something wrong,feel free to report the post .

Fortunately, ii) and iii) were already dealt with, and we were well under way with i).
It was always and only part (i) being discussed in this thread.

I was quite happy with CW's curiosity after the insulator issue. Admittedly a sidetrack.
Good that you admit that it is a sidetrack.You should have kept OP on the main track.You could have discussed about the dielectrics and polarization in the end .In any case they had little to do with the problem.

And at the end we would also have adressed CW's unhappiness (suspicion) with the discrepancy found at r=a
Why do you want to address the main issue in the end ? Why not address it in the beginning ? Why not help the OP understand what the problem asks to do ?

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But based on this, the answer I determined was:

$$\Delta V = V_{r} - V_{\infty} = V_{r}$$

$$V_{r} = \frac{1}{4\pi \epsilon_{0}} \int \frac{dq}{r}$$

And since r < a, and there is no charge within the inner sphere, dq = 0, and hence, V = 0 for r < a.

To confirm, is this sound reasoning?
That is not correct . You need to calculate potential via work done by electric field in bringing a unit charge from infinity to 'r' (r<a) .

$$V_{r} = -\int_{∞}^{r} \vec{E} \cdot \vec{dr}$$

Do you think the work done is zero ?

ehild
Homework Helper
But based on this, the answer I determined was:

$$\Delta V = V_{r} - V_{\infty} = V_{r}$$

$$V_{r} = \frac{1}{4\pi \epsilon_{0}} \int \frac{dq}{r}$$

And since r < a, and there is no charge within the inner sphere, dq = 0, and hence, V = 0 for r < a.

To confirm, is this sound reasoning?
No, your integral formula is valid when you have some charge distribution in empty free space. Now you have the domain inside the the metal sphere, and it is surrounded by the shell. And the potential is not zero on the surface of the sphere.
You certainly have learnt that the electric field is equal to the negative gradient of the potential. The potential is a continuous function. The potential difference between two points A and B is equal to the work done on a positive unit charge when it moves from A to B.
The electric field is zero in a metal in the stationary case as BvU explained. Is there any force acting on a charge if the field is zero? So what about work? Is there any potential difference between two points inside the conductor?

ehild

BvU
Homework Helper
2019 Award
Poor CW. Caught up in overadvicitis. You did good checking with your pro. And mistrusting tekst books to a decent extent isn't a bad thing either.
So: are you confident now that you understand what is going on and what is expected from you in this exercise ? There is plenty good material in the posts from our colleagues, so hardly a need to fall into repetition.

There is one little thing I would like to check with you, though: In the original post, the two, three lines before you express your suspicion: do you realize you cannot integrate qr from infinity to r ? I don't mind the casual use of r here, but the casual swapping of ##\frac{qr^{2}}{2a^{3}} ## at infinity with ##V_{\infty}##.

In the original problem you are given that V=0 at infinite separation. It helped setting a V reference scale in the answer to ii. From there on, see the last paragraph in ehild's post.

Move on to the next exercise and don't worry about Tanya and me; we'll make up when we bump into each other in another thread. She's always right. Almost.