Potential inside concentric spherical shells with non-uniform charge density

In summary, the formula for calculating the potential inside a concentric spherical shell with non-uniform charge density is V = (Q/r)*(1/4πε<sub>0</sub>)*(1/r<sub>1</sub> - 1/r<sub>2</sub>), and the charge density affects the potential by determining the overall charge distribution. The relationship between potential and electric field is E = -dV/dr, and the potential in a shell with non-uniform charge density is generally more complex than in a shell with uniform charge density. The potential can also be negative at certain distances from the center of the shell, even with a positive total charge.
  • #1
CopyOfA
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Homework Statement


We are given a two concentric spherical shells with small radius ## a ## and larger radius ## b ##. The inner and outer shells are made of conducting material and there is a volume charge density, ##\rho\left(r\right) ##, that exists between the shells,. The boundary conditions are ##\phi\left(a\right) = 0## and ##\phi\left(b\right) = V_0##.

Show that the potential for ##a<r<b## is:

##\phi\left(r\right) = \dfrac{ab}{b-a} \left[V_0 \left(\dfrac{1}{a}- \dfrac{1}{r}\right) + \int\limits_{a}^{b} \dfrac{r'^2 dr'}{\epsilon_0} \rho\left(r\right) \left(\dfrac{1}{a}-\dfrac{1}{r_<}\right)\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right) \right] ##

where ##r_< = lesser(r,r')## and ##r_> = greater(r,r')##.

Homework Equations


[/B]
##Q\left(\mathbf{r}\right) = \int\limits_{V}\rho\left(\mathbf{r}\right)dV##

##\int\limits_{S}\mathbf{E}\cdot d\mathbf{S} = \dfrac{Q_{enclosed}}{\epsilon_0}##

##\phi\left(\mathbf{r}\right) = -\int\limits_{\mathbf{r}_0}^{\mathbf{r}} \mathbf{E}\cdot d\mathbf{l}##

##\nabla^2 \phi = -\dfrac{\rho\left(\mathbf{r}\right)}{\epsilon_0}##

The Attempt at a Solution


Since the inner surface is kept at zero potential, a charge will be induced on it's surface equal to the total charge of the charge density. And thus, the charge enclosed by a Gaussian sphere of radius ##r## (##a<r<b##) will be:

##Q\left(r\right) = -\sigma_{induced}\left(a\right) + \int\limits_{V}\rho\left(r\right)dV = -4\pi \int\limits_{a}^{b} \rho\left(r\right)r^2 dr + 4\pi\int\limits_{a}^{r}\rho\left(s\right)s^2 ds##

##E\left(r\right) = \dfrac{1}{4\pi\epsilon_0 r^2}Q\left(r\right) = \dfrac{1}{\epsilon_0 r^2} \left[ -\int\limits_{a}^{b} \rho\left(t\right) t^2 dt + \int\limits_{a}^{r}\rho\left(s\right)s^2 ds \right] ##

##\phi\left(r\right) = -\int\limits_{a}^{r} E_{r}dr = \int\limits_{a}^{r} \left\{\dfrac{1}{\epsilon_0 r'^2} \left[ -\int\limits_{a}^{b} \rho\left(t\right) t^2 dt + \int\limits_{a}^{r'}\rho\left(s\right)s^2 ds \right] dr'\right\} = \int\limits_{a}^{r} \left[\dfrac{1}{r'^2}\int\limits_{a}^{b}\dfrac{\rho\left(t\right)}{\epsilon_0}t^2 dt - \dfrac{1}{\epsilon_0 r'^2}\int\limits_{a}^{r'}\rho\left(s\right)s^2 ds \right] dr'##

Taking the first sub-integral:
##\int\limits_{a}^{b}\dfrac{\rho\left(t\right)}{\epsilon_0}t^2 dt = \int\limits_{a}^{b} \left(-\nabla^2 \phi\right) t^2 dt##

In spherical coordinates, with dependence only on ##r##,
##\nabla^2 \phi = \dfrac{1}{r^2}\dfrac{d}{dr}\left(r^2 \dfrac{d\phi}{dr} \right)##

Hence,
##\int\limits_{a}^{b} \left(-\nabla^2 \phi\right) t^2 dt = -\int\limits_{a}^{b}\left[\dfrac{1}{t^2}\dfrac{d}{dt}\left(t^2 \dfrac{d\phi}{dt} \right)\right] t^2 dt = -\int\limits_{a}^{b}\dfrac{d}{dt}\left(t^2 \dfrac{d\phi}{dt} \right) dt = -t^2\dfrac{d\phi}{dt}\bigg|_{t=a}^{b}##

Plugging into larger integral,
##\int\limits_{a}^{r} \left[\dfrac{1}{r'^2}\left(-t^2\dfrac{d\phi}{dt}\bigg|_{t=a}^{b}\right)- \dfrac{1}{\epsilon_0 r'^2}\int\limits_{a}^{r'}\rho\left(s\right)s^2 ds \right] dr' = -\int\limits_{a}^{r}\dfrac{1}{r'^2}\left[b^2\dfrac{d\phi}{dt}\bigg|_{t=b} - a^2\dfrac{d\phi}{dt}\bigg|_{t=a}\right]dr' - \int\limits_{a}^{r}\dfrac{1}{r'^2}u\left(r'\right)dr'##

where ##u\left(r'\right) = \int\limits_{a}^{r'}\dfrac{\rho\left(s\right)}{\epsilon_0}s^2 ds##

I feel like this is leading nowhere though... For one, I don't know the value of the potential derivatives anywhere, much less the two surfaces. Second, I don't have an integral of the density from ##a## to ##b##. Finally, I don't have any clue where the ##r_<## and ##r_>## come into play.

Any suggestion? Am I doing some of this work wrong?
 
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  • #2
In case anyone was wondering, I did figure this out, after much struggling. I'll leave this here for the next poor soul who needs some direction on a similar problem. The approach is to use Green's function. Consider Poisson's equation:
$$\nabla^2 \phi = -\dfrac{\rho\left(r\right)}{\epsilon_0}$$
Carrying out the derivative in only the radial direction (as our charge distribution only varies radially):

$$\dfrac{1}{r^2}\dfrac{d}{dr}\left(r^2 \dfrac{d\phi}{dr}\right) = \dfrac{1}{r^2}\left(2r\dfrac{d\phi}{dr} + r^2\dfrac{d^2 \phi}{dr^2}\right) = -\dfrac{\rho\left(r\right)}{\epsilon_0}$$
Our differential equation is:

$$r^2\dfrac{d^2 \phi}{dr} + 2r \dfrac{d\phi}{dr} = -\dfrac{r^2\rho\left(r\right)}{\epsilon_0}$$
Following convention in finding the Green’s function for this setup, multiply both sides of the equation by a scalar function [itex]\psi\left(r\right)[/itex] and integrate over the region:

$$\int\limits_{a}^{b}r^2\psi\left(r\right)\dfrac{d^2 \phi}{dr} dr + \int\limits_{a}^{b}2r\psi\left(r\right) \dfrac{d\phi}{dr} dr = \int\limits_{a}^{b}-\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right) dr$$
Performing integration by parts on each integral:

$$\begin{align}r^2\psi\left(r\right)\dfrac{d\phi}{dr}\bigg|_{a}^{b} &- \left(2r\psi\left(r\right) + r^2 \dfrac{d\psi}{dr}\right)\dfrac{d\phi}{dr}\bigg|_{a}^{b} + \int\limits_{a}^{b}\phi\left(r\right)\left(2\psi\left(r\right) + 4r\dfrac{d\psi}{dr} + r^2\dfrac{d^2\psi}{dr^2} \right)dr \\
&+ 2r\psi\left(r\right)\phi\left(r\right)\bigg|_{a}^{b} - 2\int\limits_{a}^{b}\phi\left(r\right)\psi\left(r\right)dr - 2\int\limits_{a}^{b}r\phi\left(r\right)\dfrac{d\psi}{dr}dr = K \end{align}$$

where [itex]K = \int\limits_{a}^{b}-\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right)dr[/itex].
Canceling like terms:

$$2\int\limits_{a}^{b} r\phi\left(r\right)\dfrac{d\psi}{dr}dr + \int\limits_{a}^{b} r^2 \phi\left(r\right)\dfrac{d^2\psi}{dr^2} dr + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = K$$
And we get:

$$\int\limits_{a}^{b} r^2\left(\nabla^2 \psi\right) \phi\left(r\right) dr + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = K$$
Letting [itex]\nabla^2\psi\left(r\right) = -4\pi\delta\left(\mathbf{x}-\mathbf{x}'\right) = -\dfrac{4\pi}{r^2}\delta\left(r-r'\right)[/itex] the above equation is:

$$\int\limits_{a}^{b} r^2 \left( -\dfrac{4\pi}{r^2}\delta\left(r-r'\right)\right)\phi\left(r\right) dr + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = -4\pi\phi\left(r'\right) + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = K$$

Letting [itex]\psi\left(a\right) = \psi\left(b\right) = 0,[/itex] and noticing that [itex]\phi(a) = 0, \phi(b) = V_0[/itex], the result is:
$$-4\pi\phi\left(r'\right) + r^2\left[\psi\left(r\right)\dfrac{d\phi}{dr} - \phi\left(r\right)\dfrac{d\psi}{dr}\right]\bigg|_{a}^{b} = -4\pi\phi\left(r'\right) - b^2V_0 \dfrac{d\psi}{dr} = K$$

Therefore,

$$\phi\left(r’\right) = \dfrac{1}{4\pi}\left[- b^2V_0 \dfrac{d\psi}{dr} + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right)dr\right]$$
Noting that in the region [itex]r<r’, \nabla^2 \psi = 0,[/itex] try the function

$$\psi\left(r\right) = A + Br^{-1}$$
This function satisfies the above homogenous differential equation. Applying the boundary condition at [itex]a[/itex],

$$\psi\left(a\right) = 0 = A + Ba^{-1}, \Longrightarrow B = -Aa$$
$$\psi_{1}\left(r\right) = A\left(1-ar^{-1}\right)$$
Similarly, in the region [itex]r>r’, \nabla^2 \psi = 0.[/itex]

Therefore, assuming a solution and applying a boundary condition at [itex]b[/itex],

$$\psi\left(b\right) = 0 = C + Db^{-1}, \Longrightarrow D = -Ca$$
$$\psi_{2}\left(r\right) = C\left(1-br^{-1}\right)$$
In order to determine the value of the coefficients, one must use the following equations:
$$\psi_{1}\left(r’\right) = \psi_{2}\left(r’\right)$$
$$\int\limits_{r’-\varepsilon}^{r’+\varepsilon} \nabla^2 \psi dr = \int\limits_{r’-\varepsilon}^{r’+\varepsilon} -\dfrac{4\pi}{r^2}\delta\left(r-r’\right)dr = -\dfrac{4\pi}{r’^2}$$
Hence,

$$A\left(1-ar'^{-1}\right) = C\left(1-br'^{-1}\right) \Longrightarrow A = C\left(\dfrac{r'-b}{r'-a}\right)$$
and

$$\int\limits_{r’-\varepsilon}^{r’+\varepsilon} \left[\dfrac{2}{r}\dfrac{d\psi}{dr} + \dfrac{d^2\psi}{dr^2}\right]dr = \dfrac{2}{r}\psi\left(r\right)\bigg|_{r’-\varepsilon}^{r’+\varepsilon} + \int\limits_{r’-\varepsilon}^{r’+\varepsilon}\dfrac{2}{r^2}\psi\left(r\right) + \dfrac{d\psi}{dr}\bigg|_{r’-\varepsilon}^{r’+\varepsilon}$$
It's clear that the first two terms in will vanish as [itex]\varepsilon \rightarrow 0[/itex]. The third term is:

$$\dfrac{d\psi}{dr}\bigg|_{r’-\varepsilon}^{r’+\varepsilon} = \dfrac{d\psi_{2}\left(r\right)}{dr}\bigg|_{r = r' +\varepsilon} - \dfrac{d\psi_{1}\left(r\right)}{dr}\bigg|_{r = r' - \varepsilon} = Cb\left(r'+\varepsilon\right)^{-2} - Aa\left(r'-\varepsilon\right)^{-2}$$
Letting [itex]\varepsilon \rightarrow 0[/itex],

$$Cbr'^{-2} - Aar'^{-2} = -\dfrac{4\pi}{r'^2}$$
$$Cb - Aa = -4\pi$$
$$Cb - Ca\left(\dfrac{r'-b}{r'-a}\right) = -4\pi$$
$$C = -\dfrac{4\pi}{b-a}\left(\dfrac{r'-a}{r'}\right) \Longrightarrow A = -\dfrac{4\pi}{b-a}\left(\dfrac{r'-b}{r'}\right)$$
Hence,
$$\psi\left(r,r'\right) = \begin{cases} -\dfrac{4\pi}{b-a}\left(\dfrac{r'-b}{r'}\right)\left(1-ar^{-1}\right) & r<r' \\
-\dfrac{4\pi}{b-a}\left(\dfrac{r'-a}{r'}\right)\left(1-br^{-1}\right) & r>r' \end{cases}$$
$$\psi\left(r,r'\right) = \begin{cases} -\dfrac{4\pi}{b-a}\left(\dfrac{r'-b}{r'}\right)\left(\dfrac{r-a}{r}\right) & r<r' \\
-\dfrac{4\pi}{b-a}\left(\dfrac{r'-a}{r'}\right)\left(\dfrac{r-b}{r}\right) & r>r' \end{cases}$$
$$\psi\left(r,r'\right) = \begin{cases} -\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{b} - \dfrac{1}{r'}\right)\left(\dfrac{1}{a} - \dfrac{1}{r}\right) & r<r' \\
-\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)\left(\dfrac{1}{b} - \dfrac{1}{r}\right) & r>r' \end{cases}$$
Finally,
$$\psi\left(r,r'\right) = \dfrac{4\pi ab}{b-a}\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right)\left(\dfrac{1}{a} - \dfrac{1}{r_<}\right)$$
where [itex]r_< = \min\left(r,r'\right)[/itex] and [itex]r_> = \max\left(r,r'\right)[/itex].

We also need [itex]d\psi/dr[/itex] at [itex]r=b[/itex]:
$$\dfrac{d\psi}{dr} \bigg|_{r=b} = -\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)\dfrac{1}{r^2}\bigg|_{r=b} = -\dfrac{4\pi a}{b\left(b-a\right)}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)$$
Plugging this into the equation for the potential determined earlier,
$$\phi\left(r’\right) = \dfrac{1}{4\pi}\left[- b^2V_0 \dfrac{d\psi}{dr} + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\psi\left(r\right)dr\right]$$
$$\phi\left(r’\right) = \dfrac{1}{4\pi}\left[-b^2V_0\left(-\dfrac{4\pi a}{b\left(b-a\right)}\left(\dfrac{1}{a} - \dfrac{1}{r'}\right)\right) + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\dfrac{4\pi ab}{b-a}\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right)\left(\dfrac{1}{a} - \dfrac{1}{r_<}\right)dr\right]$$
$$\phi\left(r’\right) = \dfrac{ab}{b-a}\left[V_0\left(\dfrac{1}{a} - \dfrac{1}{r'}\right) + \int\limits_{a}^{b}\dfrac{r^2\rho\left(r\right)}{\epsilon_0}\left(\dfrac{1}{r_>} - \dfrac{1}{b}\right)\left(\dfrac{1}{a} - \dfrac{1}{r_<}\right)dr\right]$$
 
Last edited:

1. What is the formula for calculating the potential inside a concentric spherical shell with non-uniform charge density?

The formula for calculating the potential inside a concentric spherical shell with non-uniform charge density is V = (Q/r)*(1/4πε0)*(1/r1 - 1/r2), where Q is the total charge inside the shell, r is the distance from the center of the shell, and r1 and r2 are the inner and outer radii of the shell, respectively.

2. How does the charge density affect the potential inside a concentric spherical shell?

The charge density affects the potential inside a concentric spherical shell by determining the overall charge distribution within the shell. Non-uniform charge density can result in variations in potential at different distances from the center of the shell.

3. What is the relationship between the potential and the electric field inside a concentric spherical shell with non-uniform charge density?

The relationship between the potential and the electric field inside a concentric spherical shell with non-uniform charge density is given by E = -dV/dr, where E is the electric field and V is the potential. This means that the electric field is the negative gradient of the potential, and it is also affected by the non-uniform charge density.

4. How does the potential inside a concentric spherical shell with non-uniform charge density compare to that of a shell with uniform charge density?

The potential inside a concentric spherical shell with non-uniform charge density is generally more complex and difficult to calculate than that of a shell with uniform charge density. This is because the non-uniform charge density leads to variations in potential at different distances from the center of the shell, whereas a uniform charge density would result in a simpler, more symmetrical potential distribution.

5. Can the potential inside a concentric spherical shell with non-uniform charge density be negative?

Yes, the potential inside a concentric spherical shell with non-uniform charge density can be negative. This is because the potential is determined by the charge distribution within the shell, and a non-uniform charge density can result in regions of higher and lower potential. In fact, the potential can be negative at certain distances from the center of the shell, even if the total charge inside the shell is positive.

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