Question regarding the energy stored in a conductor

[Laudator]

A conducting spherical shell with inner radius $a$ and outer radius $b$, a point charge, $+q$, is located at the center of the cavity, inducing $-q$ on the cavity wall and $+q$ on the shell. First, I want to know the potential in different region of this configuration.

My thought:
When $r>b$, the potential behaves like the potential of a point charge, so $V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}$.
When $a<r<b$, a conductor is an equapotential, so again, $V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}$.
When $r<a$, I'm a bit confused, but I think it's $V=\frac{1}{4\pi\epsilon_0}(\frac{q}{r}+\frac{-q}{a}+\frac{q}{b})$.

Is this correct? Second, what is the energy stored in this configuration?

My thought:
I learned that the energy stored in a surface charge distribution is $E=\frac{1}{2}\int\sigma Vda$. The surface charge densities on the cavity wall and the outer shell should all be uniform and the potentials on them are the same constant, but the surface charge densities have opposite sign, does this mean the energy is 0? This doesn't seem right... Also I don't know how to take the energy of the point charge at the origin into account, the integral seems only applies to distributions, not an isolated charge.

This is, in fact, a problem in David J. Griffiths' Introduction to Electrodynamics, 4th edition, problem 2.60. I checked the solution manual, the answer is $-\frac{q^2}{8\pi\epsilon_0}(\frac{1}{a}-\frac{1}{b})$. I brainstormed for days, still don't get it...

 

[Andrew Mason]

A conducting spherical shell with inner radius $a$ and outer radius $b$, a point charge, $+q$, is located at the center of the cavity, inducing $-q$ on the cavity wall and $+q$ on the shell. First, I want to know the potential in different region of this configuration.

My thought:
When $r>b$, the potential behaves like the potential of a point charge, so $V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}$.
When $a<r<b$, a conductor is an equapotential, so again, $V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}$.
When $r<a$, I'm a bit confused, but I think it's $V=\frac{1}{4\pi\epsilon_0}(\frac{q}{r}+\frac{-q}{a}+\frac{q}{b})$.
F
Is this correct?

This is an application of Gauss' Law. What is the enclosed charge in a Gaussian sphere of radius r<a? It is a bit trickier for a<r<b. You have to know that the charge is on the surfaces and the charges arrange themselves to have 0 electric field inside the conductor.

AM

 

[Laudator]

What is the enclosed charge in a Gaussian sphere of radius r<a?

It's $+q$ and according to Gauss' Law, the field behaves like a point charge when $r<a$, so $\mathbf{\vec{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\mathbf{\vec{r}}$, that's what's confusing for me because $\mathbf{\vec{E}}=-\nabla V$, adding a constant to $V$ won't change the field. The answer given in "My thought" above gives exactly this field and gives the correct fields for $a<r<b$ and $r>b$ too. Does this mean I'm right? You didn't give me an answer. Also, what about the second part? The energy stored?

 

[sophiecentaur]

Would it be over-trivialising the problem to suggest that the outer sphere can be treated as it it's not there when calculating the potential where a<r<b ? If the field inside b, due to charge on b is zero, can't b just be ignored?

 

[Laudator]

Would it be over-trivialising the problem to suggest that the outer sphere can be treated as it it's not there when calculating the potential where a<r<b ? If the field inside b, due to charge on b is zero, can't b just be ignored?

I don't understand what you are suggesting.

 

[Andrew Mason]

It's $+q$ and according to Gauss' Law, the field behaves like a point charge when $r<a$, so $\mathbf{\vec{E}}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\mathbf{\vec{r}}$, that's what's confusing for me because $\mathbf{\vec{E}}=-\nabla V$, adding a constant to $V$ won't change the field. The answer given in "My thought" above gives exactly this field and gives the correct fields for $a<r<b$ and $r>b$ too. Does this mean I'm right? You didn't give me an answer.

Your answers are correct. But since you are not sure why you are right, your reasoning needs to be clarified. Can you explain how you got the answer for r<a? Try starting with the definition of electric potential in terms of the work done on or by a unit charge in moving from its present position to a position at infinity. If you do that and then add the work done or required in moving from that position to another, you will be able to explain your answer and you will know why it is correct.

Also, what about the second part? The energy stored?

Once we are clear on the physics, we can work out the energy. Think of the energy of the configuration as the energy required to move the conducting shell there from a very far distance away. (Hint: apply Gauss' law to a conducting sphere of the same thickness but with a radius >>>b and then shrink it down to the present configuration. Does surface charge density matter?)

AM

 

[Laudator]

Can you explain how you got the answer for r<a?

First, I applied Gauss' Law for $r>b$, the total charge inside is $(+q)_\text{point charge at O}+(-q)_\text{charge on sphere a}+(+q)_\text{charge on sphere b}=q$, so the field behaves like a point charge. The potential of a point charge is given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$ and when $r=b$, the potential is thus given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}$. Next, because I know a conductor is an equipotential, the potential at $a$ must be equal to this value and finally, after applying Gauss' Law to $r<a$ and knowing also that the potential is continuous across any surface charge distribution, I reach the conclusion that the potential within should be $V=\frac{1}{4\pi\epsilon_0}(\frac{q}{r}-\frac{q}{a}+\frac{q}{b})$.

This is my train of thought. I will use your suggestion, resorting back to the basics and use work done by a unit charge to work this out again, but at least now I know I have this 1st part of the question figured out.

(Hint: apply Gauss' law to a conducting sphere of the same thickness but with a radius >>>b and then shrink it down to the present configuration. Does surface charge density matter?)

As for your hint, I'm not sure how to use it... I don't know how to relate Gauss' Law (which deals with the field) to Energy. Are you hinting at this formula? $$W=\frac{\epsilon_0}{2}\int E^2d\tau$$ Also, the reason I want to know the potential is that I want to figure this out using $$E=\frac{1}{2}\int\sigma Vda$$ Is it possible to figure this out from the potential alone?

 

[sophiecentaur]

I don't understand what you are suggesting.

Hmm, I think your query is justified. I was confusing Field with Potential

 

[Andrew Mason]

First, I applied Gauss' Law for $r>b$, the total charge inside is $(+q)_\text{point charge at O}+(-q)_\text{charge on sphere a}+(+q)_\text{charge on sphere b}=q$, so the field behaves like a point charge. The potential of a point charge is given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$ and when $r=b$, the potential is thus given by $V=\frac{1}{4\pi\epsilon_0}\frac{q}{b}$. Next, because I know a conductor is an equipotential, the potential at $a$ must be equal to this value and finally, after applying Gauss' Law to $r<a$ and knowing also that the potential is continuous across any surface charge distribution, I reach the conclusion that the potential within should be $V=\frac{1}{4\pi\epsilon_0}(\frac{q}{r}-\frac{q}{a}+\frac{q}{b})$.

This is my train of thought. I will use your suggestion, resorting back to the basics and use work done by a unit charge to work this out again, but at least now I know I have this 1st part of the question figured out.

As long as you understand that the potential at r<a is found by integrating $\vec{E} \cdot d\vec{s}$ field over a path from a to r<a and adding the potential at a<r<b ie. the path integral of the field E from infinity to a.

As for your hint, I'm not sure how to use it... I don't know how to relate Gauss' Law (which deals with the field) to Energy. Are you hinting at this formula? $$W=\frac{\epsilon_0}{2}\int E^2d\tau$$ Also, the reason I want to know the potential is that I want to figure this out using $$E=\frac{1}{2}\int\sigma Vda$$ Is it possible to figure this out from the potential alone?

You have to use the given potential energy of the charge distribution and add it to the energy of moving that charge distribution from infinity to the position in question with respect to the charge at the origin. Be careful with the signs when adding. (Edit Note: You should check your equation for the energy of a charge distribution. It is a double integral because one has to sum the potential of each charge with respect to the others and then sum for all the charges (and since you are counting the potentials of each charge twice, divide by 2). So the equation for the energy of a charge distribution is (using U for potential energy rather than E to avoid confusion with E as the electric field) $U=\frac{1}{2}\int\int\sigma Vda$

AM

 

[vanhees71]

You have (in electrostatics)
$$\vec{E}=-\vec{\nabla} V, \quad \vec{\nabla} \cdot \vec{E}=\frac{1}{\epsilon_0} \rho.$$
Now you can use
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{E}^2(x)=-\int_{\mathbb{R}^3} \mathrm{d}^3 x \vec{E} \cdot \vec{\nabla} V= + \int_{\mathbb{R}^3} \mathrm{d}^3 x V \vec{\nabla} \cdot \vec{E} = \frac{1}{\epsilon_0} \int_{\mathbb{R}^3} \mathrm{d}^3 x \rho V.$$
Now if you have surface charges instead of the volume integral you have a surface integral as you've written.

 

[Laudator]

@Andrew Mason @vanhees71 While calculating the energy stored in a charged sphere, I read from here. The author started from a sphere which is already charged with $q$ and then bring $dq$ from infinity onto the sphere with radius $a$ and the work it takes is:$$W=\int^a_\infty\vec F\cdot d\vec x=\int^a_\infty dq\vec E\cdot d\vec x=-\frac{qdq}{4\pi\epsilon_0}\int^a_\infty\frac{dr}{r^2}$$Then he proceeds to integrate $dq$. Here's what I got confused, it looks awfully like the work it takes to bring a point charge $dq$ from infinty to $a$ without smearing it uniformly onto the sphere. The smearing takes no work?

 

[vanhees71]

No, because there's no force perpendicular to the sphere (the original charge distribution is spherically symmetric, as seems to be assumed since the author calculates with a Coulomb field of charge $q$).

 

[Laudator]

No, because there's no force perpendicular to the sphere (the original charge distribution is spherically symmetric, as seems to be assumed since the author calculates with a Coulomb field of charge $q$).

I suppose you meant "tangent to"?

 

[Laudator]

I think I got the second part of my question worked out, thanks to @vanhees71, now I know the smearing of $dq$ takes no work. Here's my train of thought:

I will treat the conductor as two conducting spheres with nothing in between. Originally, there's only a point charge at the origin and the field it creates behaves like $\vec E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat r$. Next, I'll bring in a point charge $dq$ from infinity to $r=a$, which takes work:$$dW=\int^a_\infty dq\vec E\cdot d\vec r=-\frac{qdq}{4\pi\epsilon_0}\int^a_\infty\frac{dr}{r^2}=\frac{qdq}{4\pi\epsilon_0}\left.\frac{1}{r}\right|^a_\infty=\frac{qdq}{4\pi\epsilon_0a}$$Then smear it over the sphere of radius $a$, this takes no work as the process involves only moving along a path that is always perpendicular to the force, $\vec F\cdot d\vec x=0$. Integrating $dq$ and as the inner cavity wall contains negative charge, the limit of integration goes from $+q$ to $0$. So $$W=\int^0_q dW=\int^0_q \frac{qdq}{4\pi\epsilon_0a}=-\frac{q^2}{8\pi\epsilon_0a}$$Next, repeat the process, but bring $dq$ to $r=b$, the work it takes is $dW=\frac{qdq}{4\pi\epsilon_0b}$. Integrate $dq$, the limit of integration this time is from $0$ to $+q$. So $$W=\int^q_0 dW=\int^q_0 \frac{qdq}{4\pi\epsilon_0b}=\frac{q^2}{8\pi\epsilon_0b}$$ And thus the total energy stored in the configuration is$$W=-\frac{q^2}{8\pi\epsilon_0}(\frac{1}{a}-\frac{1}{b})$$Same as the answer given by David J. Griffiths' book. However, I can't help but feel finding the potentials in different region of this configuration useless (the problem didn't ask that, it's just I thought it's useful...).

 

[Andrew Mason]

... So $$W=\int^q_0 dW=\int^q_0 \frac{qdq}{4\pi\epsilon_0b}=\frac{q^2}{8\pi\epsilon_0b}$$ And thus the total energy stored in the configuration is$$W=-\frac{q^2}{8\pi\epsilon_0}(\frac{1}{a}-\frac{1}{b})$$Same as the answer given by David J. Griffiths' book. However, I can't help but feel finding the potentials in different region of this configuration useless (the problem didn't ask that, it's just I thought it's useful...).

That gives you the self-energy of the charge distribution without taking into account the central charge. Now add in the potential energy of the charges on the conducting sphere with respect to the central charge from the potentials that you found in the first part.

$$W=-\frac{q^2}{8\pi\epsilon_0}(\frac{1}{a}-\frac{1}{b}) + q\left(-\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r}+\frac{-q}{a}+\frac{q}{b}\right)\right)$$ (for the potential of the inner sphere surface r=a)

AM

 

[Laudator]

That gives you the self-energy of the charge distribution without taking into account the central charge. Now add in the potential energy of the charges on the conducting sphere with respect to the central charge from the potentials that you found in the first part.

$$W=-\frac{q^2}{8\pi\epsilon_0}(\frac{1}{a}-\frac{1}{b}) + q\left(-\frac{1}{4\pi\epsilon_0}\left(\frac{q}{r}+\frac{-q}{a}+\frac{q}{b}\right)\right)$$ (for the potential of the inner sphere surface r=a)

AM

Em... I'm only getting more confused... The problem in the book asked the total energy stored in the configuration, the answer is the one I mentioned, the author didn't take the central point charge into account. He stated that the first step of forming this configuration is to move in the central charge, but there's no field in the beginning, so it takes no work. What is the answer you just provided then? What does that work represent? Also, why is there's a minus sign in front of $\frac{1}{4\pi\epsilon_0}$? The one I calculated in the first part is positive.

 

[vanhees71]

I suppose you meant "tangent to"?

Sure ;-)).

 

[Andrew Mason]

Em... I'm only getting more confused... The problem in the book asked the total energy stored in the configuration, the answer is the one I mentioned, the author didn't take the central point charge into account. He stated that the first step of forming this configuration is to move in the central charge, but there's no field in the beginning, so it takes no work. What is the answer you just provided then? What does that work represent? Also, why is there's a minus sign in front of $\frac{1}{4\pi\epsilon_0}$? The one I calculated in the first part is positive.

Sorry to cause confusion! I was just looking at the inner surface of the conducting shell (r=a and charge = -q). The potential energy of both shell surfaces (ie. the whole shell) with respect to the central charge is 0 because the potential with respect to the central charge is the same for both surfaces (r=a and r=b) and the total charge is 0.

The energy of the charge distribution itself is a bit more subtle. If you take away the central charge, there is no net charge so the energy is 0. But when the central charge is inserted the charges distributed throughout the conducting sphere m9ve to the outer surfaces. This involves a change in energy in separating the charges of the conducting sphere.

The charge on each surface is created by half the charges moving to surface a an average distance of (b-a)/2 in the direction of $-\vec{r}$ and the other half (opposite sign) moving to surface b an average distance of (b-a)/2 in the +$\vec{r}$ direction. They do that under the influence of the field of the enclosed charge. During this time, the field between a and b decreases with each charge that moves to surface a. The work done could be determined.

for example, for surface a:

$$W_a = \int_0^{-q}dq\int_{a + (b-a)/2}^a \vec{E}\cdot d\vec{r}$$

If we let the opposite charges move to surface a first, then between the two surfaces the enclosed charge = 0 so E is zero and $$W_b = 0$$

Anyhow, that seems to me to be what the energy consists of. Perhaps others may wish to comment.

AM