Minimizing the Energy of Two Conductors Very Far Apart

[zenterix]

Homework Statement

Consider two metal shells of radius $R_1$ and $R_2$. Assume they are very far apart so that the potential near one shell depends only on the charge on that shell. Assume that because of this large distance, any charge on a shell distributes itself uniformly.

a) Assume we place $Q-q$ on shell 1 and $q$ on shell 2. What is the total potential energy of the two shells with this arrangement of charges?

b) What is the value of $q$ that minimizes total energy?

c) Given the distribution of charge which minimizes the total energy, what is the potential difference between the shells?

Relevant Equations

The electric field of a spherical shell with radius $r$ and charge $Q$ at a point on the outside of the shell is $\frac{Q}{4\pi\epsilon_0 r^2}\hat{r}$.

The work done by the electric field when we bring a charge $dq$ from an infinite distance to the surface of a shell with radius $r$ is

$$dW=\int_{\infty}^r \frac{Qdq}{4\pi\epsilon_0 r^2}dr=-\frac{Qdq}{4\pi\epsilon_0r}\tag{1}$$

The work done by the electric field to charge a spherical shell from charge $0$ to $Q$, bringing charge from an infinite distance away is

$$W=-\int_0^Q \frac{qdq}{4\pi\epsilon_0 r}=-\frac{Q^2}{8\pi\epsilon_0 r}\tag{2}$$

The potential energy of such a charged spherical shell is $U=-W$.

We can apply these results to give us the potential energy of each of our spherical shells.

The potential energy of shell 1 is

$$U_1=\frac{(Q-q)^2}{8\pi\epsilon_0 R_1}\tag{3}$$

and the potential energy of shell 2 is

$$U_2=\frac{q^2}{8\pi\epsilon_0 R_2}\tag{4}$$

The potential energy of the system is the sum of the individual potential energies

$$U=\frac{1}{8\pi\epsilon_0}\left ( \frac{(Q-q)^2}{R_1}+\frac{q^2}{R_2} \right )\tag{5}$$

$$=\frac{1}{8\pi\epsilon_0}\left ( \frac{q^2R_1+(Q-q)^2R_2}{R_1R_2} \right )\tag{6}$$

The above result answers part (a).

Considering this as a function of $q$, we differentiate and equate to zero to find the minimum of $U$, which is at

$$q=\frac{QR_2}{R_1+R_2}\tag{7}$$

The second derivative $U''$ is larger than zero so we do indeed have a minimum. (This answers part (b))

We can plug (7) into (6) to find that the minimum of $U$ is

$$U_{\text{min}}=\frac{1}{8\pi\epsilon_0}\frac{Q^2}{R_1+R_2}\tag{8}$$

This problem is from MIT Open Learning Library and I can input my answers and check them. The above has all checked out.

My question is about part (c).

I know the answer is that the potential difference between the shells is zero. The reason I know is because for some reason I mistakenly thougth I had found this answer in my equations, I gave this answer to MIT OLL and got it right.

But then I realized that I actually hadn't found the answer in my equations. I simply got lucky.

I am a bit confused about how to show that the shells are at the same potential when we minimize the system potential energy.

Intuitively, it seems to make sense because as we saw above, there is relatively more charge on shell 2 but the larger radius means lower potential at the surface.

There is less charge on shell 1 but higher potential at its surface.

However, when I go to the equations I find

$$U_1=\frac{1}{8\pi\epsilon_0}\frac{(Q-q)^2}{R_1}$$
$$=\frac{1}{8\pi\epsilon_0} \frac{Q^2 R_1^2}{(R_1+R_2)^2}\frac{1}{R_1}$$
$$=\frac{1}{8\pi\epsilon_0} \frac{Q^2R_1}{(R_1+R_2)^2}\tag{9}$$

and

$$U_2=\frac{1}{8\pi\epsilon_0}\frac{q^2}{R_2}$$
$$=\frac{1}{8\pi\epsilon_0} \frac{Q^2 R_2^2}{(R_1+R_2)^2}\frac{1}{R_2}$$
$$=\frac{1}{8\pi\epsilon_0} \frac{Q^2R_2}{(R_1+R_2)^2}\tag{10}$$

These are the potential energies of the shells.

Now, it seems that if I divide (9) by $Q-q=\frac{QR_1}{R_1+R_2}$ and divide (10) by $q=\frac{QR_2}{R_1+R_2}$ then I get equality. But this is just an observation. It is not clear to me, in this problem, why I would (be allowed to) do that.

I mean, I sort of see that potential difference is potential energy per unit charge, and so if I divide the potential energies I should get potential difference.

And I guess that is probably the answer to (c).

But then I did another calculation to compute the potential difference directly.

For shell 1

$$\int_{R_1}^{d-R_2}\frac{(Q-q)}{4\pi\epsilon_0 r^2}dr$$

$$=\frac{Q(d-R_1-R_2)}{4\pi\epsilon_0(R_1+R_2)(d-R_2)}\tag{11}$$

and for shell 2

$$\int_{d-R_1}^{R_2}\frac{q}{4\pi\epsilon_0 r^2}dr$$

$$=\frac{-Q(d-R_1-R_2)}{4\pi\epsilon_0(R_1+R_2)(d-R_1)}\tag{12}$$

(11) and (12) have opposite signs and their magnitude differs by a factor in the denominator.

Either my calculations are wrong or I seem to be missing a very small step to make these equal.

Shouldn't (11) and (12) sum to zero?

 

[anuttarasammyak]

I have not investigated it in detail, but you applied approximation
$$d >> R_1,R_2$$
to get the solution. You do not have to be sensitive to the difference of $d-R_1$ and $d-R_2$.

 

[Delta2]

Homework Statement: Consider two metal shells of radius $R_1$ and $R_2$. Assume they are very far apart so that the potential near one shell depends only on the charge on that shell. Assume that because of this large distance, any charge on a shell distributes itself uniformly.

a) Assume we place $Q-q$ on shell 1 and $q$ on shell 2. What is the total potential energy of the two shells with this arrangement of charges?

b) What is the value of $q$ that minimizes total energy?

c) Given the distribution of charge which minimizes the total energy, what is the potential difference between the shells?
Relevant Equations: The electric field of a spherical shell with radius $r$ and charge $Q$ at a point on the outside of the shell is $\frac{Q}{4\pi\epsilon_0 r^2}\hat{r}$.

And I guess that is probably the answer to (c).

Up to this line all seem good to me. Just to emphasize that the potential at the surface of a spherical shell, equal its (self) potential energy divided by its total charge (there is also a factor of 2 in front I think, that is potential $V=2U/Q$). (What you calculate is the so called self-potential energy of each shell , that is the potential energy of its charge due to the field of this charge.)

But then after this line I seriously don't understand what you doing when you go to calculate the potential difference in a different way.

Seems to me you calculate the line integral of the Electric field due to the one shell from the surface of this shell till the surface of the other shell, this is not the potential at the surface of the shell (neither the potential difference between the two shells), you got to take this integral from the surface of the shell till infinity.

The problem in this problem is that it is asking you to treat each shell like it is isolated, when you calculate each shell's potential energy but then it creates a confusion because it asks for the potential difference between the surfaces of each shell. And you fall victim to this confusion when you try to compute the potential difference in the alternative way.

 

[haruspex]

These are the potential energies of the shells.

Yes, but you are asked for the difference in potential, not the difference in potential energies.

 

[zenterix]

Seems to me you calculate the line integral of the Electric field due to the one shell from the surface of this shell till the surface of the other shell, this is not the potential at the surface of the shell (neither the potential difference between the two shells), you got to take this integral from the surface of the shell till infinity.

Isn't potential difference between two arbitrary points in space computed as

$$\Delta V=-\int_C \vec{E}\cdot d\vec{r}$$

That, is the line integral of the electric field on a curve $C$ between the two points?

I used a line integral between a point $A$ on one shell and a point $B$ on the other shell.

I computed two line integrals, one for each individual electric field.

I then hoped that adding them would result in zero.

On the other hand if we just look at the electric fields due to each shell we already see that they are different.

For a point on the line connecting the centers of the shells where $r$ is the distance from the center of shell 1, we have

$$E_1=\frac{QR_1}{4\pi\epsilon_0(R_1+R_2) r^2}$$

$$E_2=\frac{QR_2}{4\pi\epsilon_0(R_1+R_2) (d-r)^2}$$

 

[zenterix]

I have not investigated it in detail, but you applied approximation
$$d >> R_1,R_2$$
to get the solution. You do not have to be sensitive to the difference of $d-R_1$ and $d-R_2$.

I agree. Nonetheless, if I do the calculations exactly, shouldn't (11) be equal to (12)?

Using the approximation these expressions are equal. But why aren't they equal without the approximation?

 

[zenterix]

Yes, but you are asked for the difference in potential, not the difference in potential energies.

Indeed, that is why I continued on with the calculation and divided by the charges to get the potential differences between the shell surfaces and infinity, and then subtracted one potential difference from the other to obtain zero.

However, I would like to know why I can't do the calculation of potential difference directly between points on the surfaces.

 

[zenterix]

I was able to obtain the expected result by taking into account the effect of one shell on the charging of the other.

For shell 1, to bring a charge $dq$ from a radius of $\infty$ to a radius of $r_1$ the work done by the electric field of shell 1 is

$$\int_{\infty}^{r_1} \frac{Qdq}{4\pi\epsilon_0r^2}dr=-\frac{Qdq}{4\pi\epsilon_0r_1}$$

To charge up shell 1 from zero charge to a charge $Q-q$ the work done by the electric field is then

$$\int_0^{Q-q}\left ( -\frac{q}{4\pi\epsilon_0r_1} \right )dq$$

$$=-\frac{(Q-q)^2}{8\pi\epsilon_0r_1}$$

Similarly, to charge up shell 2 from zero charge to charge $q$, the work done by the electric field of shell 2 is

$$-\frac{q^2}{8\pi\epsilon_0r_2}$$

When we are charging up shell 2, if we don't disregard the effect of the electric field of shell 1 then we have extra work done. We calculate this extra work as follows.

For each $dq$ we add to shell 2, the work done by shell 1's electric field is

$$\int_\infty^d \frac{(Q-q)dq}{4\pi\epsilon_0r^2}dr=-\frac{(Q-q)dq}{4\pi\epsilon_0 d}$$

(I used $d$ here because on average this is the distance from shell 1)

and the total work is

$$(Q-q)\int_0^q \left ( -\frac{1}{4\pi\epsilon_0 d} \right )dq$$

$$=-\frac{(Q-q)q}{4\pi\epsilon_0 d}$$

The total potential energy of the system is then the sum of the negative of these three work calculations

$$U=\frac{(Q-q)^2}{8\pi\epsilon_0r_1}+\frac{q^2}{8\pi\epsilon_0r_2}+\frac{(Q-q)q}{4\pi\epsilon_0 d}$$

If we differentiate $U$ with respect to $q$, equate to zero, and solve for $q$ we reach

$$q=\frac{Qr_2(d-r_1)}{dr_1+dr_2-2r_1r_2}$$

which is the charge on shell 2.

The charge on shell 1 is $Q-q$ which is

$$\frac{Qr_1(d-r_2)}{dr_1+dr_2-2r_1r_2}$$

At this point, we know the electric fields due to each shell and we can try to calculate the potential difference between points on the shells (along the straight line connecting their centers).

The potential difference due to shell 1's electric field between a point on shell 1 and a point on shell 2 is.

$$\int_{r_1}^{d-r_2} \frac{Qr_1(d-r_2)}{dr_1+dr_2-2r_1r_2}\frac{1}{4\pi\epsilon_0}\frac{1}{r^2}dr$$

$$=\frac{Q(d-r_2-r_1)}{4\pi\epsilon_0(dr_1+dr_2-2r_1r_2)}\tag{1}$$

The same calculation for the electric field of shell 2 gives

$$\int_{d-r_1}^{r_2}\frac{Qr_2(d-r_1)}{dr_1+dr_2-2r_1r_2}\frac{1}{4\pi\epsilon_0}\frac{1}{r^2}dr$$

$$=-\frac{Q(d-r_2-r_1)}{4\pi\epsilon_0(dr_1+dr_2-2r_1r_2)}\tag{2}$$

(1) and (2) cancel, so the potential difference is zero.

 

[haruspex]

So your problem was that you correctly ignored the term of order $1/d$ in part a) but not in part c)?

 

[Delta2]

What you do at post #8 is correct but it is not within the allowed framework set by the problem statement. According to problem statement:
"Assume they are very far apart so that the potential near one shell depends only on the charge on that shell"
Don't you think that at #8 you violate this assumption? Both when you calculate the potential energy, which is no longer the self potential energy, and when you calculate the potential difference.

 

[anuttarasammyak]

I agree. Nonetheless, if I do the calculations exactly, shouldn't (11) be equal to (12)?

In exact solution including order of R/d and higher, interaction between the spheres should be considered. Charge distribution becomes inhomogeneous. I wonder whether the approximate solution you get should show the equality.

 

[Delta2]

Hmm, is it something obvious that minimum of the total potential energy leads to equal potentials in the surface of each sphere?

 

[haruspex]

Hmm, is it something obvious that minimum of the total potential energy leads to equal potentials in the surface of each sphere?

Yes, it is obvious.

 

[Delta2]

Yes, it is obvious.

Not obvious to me!

 

[Delta2]

If I am to prove it I was going to use the $V=2\frac{U}{Q}$ equation for each sphere and then do some algebra and calculus, that doesn't seem so obvious to me.

 

[haruspex]

Not obvious to me!

Assuming the charges are positive, consider moving one tiny bit of charge from the higher potential to the lower. What is the energy change?

 

[Delta2]

$\Delta V dq$

 

[Delta2]

Sorry cant see where this is leading to, neither what minimization of potential energy has to do with moving charge from one sphere to the other.
This shows that when potentials are equal the total potential energy is lower than the initial configuration. But why at minimum?

 

[Delta2]

Ok ok, we take the minimum potential energy as initial configuration and if the potentials aren't equal we can find a lower potential energy configuration where the potentials are equal. Nice proof by contradiction, it is kind of sort but not so obvious if you ask me.

I would classify this as Clever shortcut explanation where the clever shortcut is indeed a shortcut but not an obvious one.

 

[zenterix]

@Delta2 I didn't include a part d), but in fact in the MIT course I am following, item d has to do with your question.

That item poses a similar scenario to the original one I posted, but now the shells have the same radius $R$ but one shell has charge $Q$ and the other $2Q$, as below

We connect a wire between the spheres. What happens?

The electric potential on the surfaces of the spheres are different. When the wire connects them, charge flows from the higher potential (shell 1) to the lower potential (shell 2) until the potential is the same on the two spheres.

When a positive charge flows, it flows in the direction of the electric field and so the electric work is positive. Potential energy goes down. It keeps going down until there is no more potential difference, and hence no flow of charge.

Any potential difference offers an opportunity for charge to be redistributed from a higher to a lower potential, thus lowering the potential energy of the system.

I also don't think the word "obvious" should be used to describe understanding this though.

 

[zenterix]

So your problem was that you correctly ignored the term of order $1/d$ in part a) but not in part c)?

Indeed.

As far as I can tell, with the approximation of ignoring that term, the way we can tell that the potential difference is zero is either

1) looking at (11) and (12) and realizing that $d-R_1\approx d-R_2$ thus making the sum of (11) and (12) zero.

or

2) looking at the potential energy of each shell, dividing by its charge, noting that this gives what I interpret as "average potential energy per unit charge" for each shell and noting that these values are equal.

 

[haruspex]

I also don't think the word "obvious" should be used to describe understanding this though.

I was just trying to goad @Delta2 into thinking it through.

 

[Delta2]

In science we shouldn't ask if it is obvious or not, that is subjective, but maybe we should ask if it is a short or long explanation, that isn't subjective.

 

[anuttarasammyak]

Assuming the charges are positive, consider moving one tiny bit of charge from the higher potential to the lower. What is the energy change?

In other words.
Connect the spheres with wire. Charge flows from high potential one to low potential one until they become equal potential. Therefore charge distribution for minimum energy state is accomplished.