Electrostatic energy of concentric shells

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Homework Statement

What's the electrostatic energy associated with two concentric shells with radii $a$ and $b$, $b > a$? The outer shell has a charge of $-q$, the inner shell's charge is $q$

Relevant Equations

$W = \frac{\epsilon_{0}}{2}\int |\vec{E}|² d \tau$

$W = \frac{1}{2} \int \rho V d \tau$

I know the energy is $\frac{q²}{ 8 \pi \epsilon_{0}}( \frac{1}{a} - \frac{1}{b})$, but I can't get this result using the second equation.
What I did:

$W = \frac{1}{2} \int \rho V d \tau$

$\rho = \frac{q}{ \frac{4}{3} \pi r³}, a < r < b$

$V = \frac{q}{4 \pi \epsilon_{0} r}$

$W = \frac{ 3 q²} {16 \pi² \epsilon_{0}} \int 4 \pi \frac {r²}{r⁴} dr = \frac{3 q²}{ 8 \pi \epsilon_{0}}( \frac{1}{a} - \frac{1}{b})$

I didn't get rid of that factor of 3, where did I go wrong?

 

[kuruman]

This is a spherical capacitor set up. What is the energy stored in a capacitor charged to q?

 

[Steve4Physics]

Using equations with $\rho$ makes no sense here.

$\rho$ is the charge density for charge spread throughout a 3D volume (a simlple example is solid insulating sphere with the charge uniformly distributed throughout the sphere).

$\rho$ is irrelevant here because we have have charges distributed on (infinitely thin) spherical shells, with nothing in-between the shells.

Your first equation $W = \frac{\epsilon_{0}}{2}\int |\vec{E}|² d \tau$ might be more useful because it gives the energy stored in the electric field (in the space between the shells), which is is what is required.

 

[etotheipi]

Using equations with ρ makes no sense here.

ρ is the charge density for charge spread throughout a 3D volume (a simlple example is solid insulating sphere with the charge uniformly distributed throughout the sphere).

ρ is irrelevant here because we have have charges distributed on (infinitely thin) spherical shells, with nothing in-between the shells.

`I agree it's a bad approach to try working with $\rho$, but we can define a perfectly workable definition of $\rho$ for an infinitely thin shell (or in this case, two shells) by way of the Dirac delta,$$\rho(r) = \frac{q}{4\pi a^2} \delta(r - a) - \frac{q}{4\pi b^2} \delta(r - b)$$

 

[haruspex]

This is a spherical capacitor set up. What is the energy stored in a capacitor charged to q?

Using the formula for capacitance of a spherical capacitor seems a bit of a cheat. Then again, doing the double area integral over the shells is overkill.
A third way is to calculate the potential at each shell and think about work done as charge leaks away to infinity from each at the same rate.

 

[Steve4Physics]

`I agree it's a bad approach to try working with $\rho$, but we can define a perfectly workable definition of $\rho$ for an infinitely thin shell by way of the Dirac delta,$$\rho(r) = \frac{q}{4\pi a^2} \delta(r - a) - \frac{q}{4\pi b^2} \delta(r - b)$$

True. I said what I said because the OP was mis-using $\rho$ and clearly not understanding its interpretation and (conventional )use. I just wanted to emphasise that (at the expense of some rigour as you point out!).

 

[etotheipi]

True. I said what I said because the OP was mis-using $\rho$ and clearly not understanding its interpretation and (conventional )use. I just wanted to emphasise that (at the expense of some rigour as you point out!).

Yeah, I suspected this was the case and agree with what you were telling the OP. I just think delta functions are cool, and couldn't resist writing it out

 

[Steve4Physics]

Yeah, I suspected this was the case and agree with what you were telling the OP. I just think delta functions are cool, and couldn't resist writing it out

Lol

 

[TSny]

For a conductor, the potential $V$ is constant. So, when doing the integral $\frac{1}{2} \int \rho V d \tau$ for one of the conducting shells, you can pull $V$ out of the integral. I don't think it matters that the charge is distributed on the surface of the conductor rather than throughout some volume of the conductor. You can always imagine in this case that the surface charge has an infinitesimal thickness so that it can be treated as a volume charge distribution. Or, you can replace $\frac{1}{2} \int \rho V d \tau$ by a surface integral $\frac{1}{2} \int \sigma V d A$.

 

[Steve4Physics]

For a conductor, the potential $V$ is constant. So, when doing the integral $\frac{1}{2} \int \rho V d \tau$ for one of the conducting shells, you can pull $V$ out of the integral. I don't think it matters that the charge is distributed on the surface of the conductor rather than throughout some volume of the conductor. You can always imagine in this case that the surface charge has an infinitesimal thickness so that it can be treated as a volume charge distribution. Or, you can replace $\frac{1}{2} \int \rho V d \tau$ by a surface integral $\frac{1}{2} \int \sigma V d A$.

Yes, etotheipi has already explained how to do this in post #4. (And see my reply in post #6.)

 

[TSny]

Yes, etotheipi has already explained how to do this in post #4. (And see my reply in post #6.)

OK.

Once you pull the potential $V$ out of the integral, then the remaining integral is just the charge on the conductor (without worrying how the charge is distributed on the conductor). I think maybe the OP was not seeing that. So, using $\frac12 \int \rho V d\tau$ is a nice way to get to the answer. It quickly yields the general capacitor formula $U = \frac 12 Q \Delta V$.

 

[etotheipi]

So, using $\frac12 \int \rho V d\tau$ is a nice way to get to the answer. It quickly yields the general capacitor formula $U = \frac 12 Q \Delta V$.

Yeah, viz$$\begin{align*}U &= \frac{1}{2}\phi_a \int_{\mathbb{R}^3} \frac{q}{4\pi a^2} \delta(r-a) dV - \frac{1}{2} \phi_b \int_{\mathbb{R}^3} \frac{q}{4\pi b^2} \delta(r-b) dV \\

&= \frac{1}{2}q(\phi_a - \phi_b)\end{align*}$$