Electrostatic energy of concentric shells

  • #1

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?
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
9,472
2,791
This is a spherical capacitor set up. What is the energy stored in a capacitor charged to q?
 
  • #3
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.
 
  • #4
etotheipi
Gold Member
2019 Award
2,704
1,623
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,$$\rho(r) = \frac{q}{4\pi a^2} \delta(r - a) - \frac{q}{4\pi b^2} \delta(r - b)$$
 
  • Like
Likes Steve4Physics
  • #5
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
34,576
5,979
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.
 
  • #6
`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!).
 
  • Like
Likes etotheipi
  • #7
etotheipi
Gold Member
2019 Award
2,704
1,623
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 😜
 
  • Like
Likes Steve4Physics
  • #8
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
 
  • #9
TSny
Homework Helper
Gold Member
12,792
3,150
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##.
 
  • #10
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.)
 
  • #11
TSny
Homework Helper
Gold Member
12,792
3,150
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##.
 
  • Like
Likes Steve4Physics and etotheipi
  • #12
etotheipi
Gold Member
2019 Award
2,704
1,623
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*}$$
 
Last edited:
  • Like
Likes TSny

Related Threads on Electrostatic energy of concentric shells

Replies
0
Views
2K
Replies
3
Views
11K
Replies
1
Views
3K
Replies
7
Views
10K
  • Last Post
3
Replies
64
Views
6K
Replies
1
Views
1K
Replies
11
Views
869
Replies
3
Views
587
Top