Electrostatics - given V, find rho and Q

[Rick16]

TL;DR: Griffiths, problem 2.46

Here is the problem statement:

The electric potential of some configuration is given by the expression $V(r) = A \frac {e^{-\lambda r}} {r}$, where A and lambda are constants. Find the electric field E(r), the charge density $\rho(r)$, and the total charge Q.

I have Griffiths' solution here. For the electric field he finds $E=Ae^{-\lambda r}(1+\lambda r)\frac {\hat r}{r^2}$. When he then takes the divergence to find rho, the expression for the divergence includes the term $\nabla\cdot (\frac {\hat r}{r^2})$, which, according to an earlier result in the text equals $4\pi\delta^3(r)$, and his complete result for rho is then $\rho = \epsilon_0 A[4\pi\delta^3(r)-\frac {\lambda^2}{r}e^{-\lambda r}]$. Next he finds that $Q=0$.

Before looking at the solution, I tried to solve this problem on my own. I got the same electric field, except that I wrote it as $E=A(\frac {\lambda e^{-\lambda r}}{r}+\frac {e^{-\lambda r}}{r^2})\hat r$. Taking the divergence I get $\rho = -\epsilon_0 A \frac {\lambda^2}{r} e^{-\lambda r}$, the same result as Griffiths', except without the delta term. Given this rho, Q is not equal to zero.

I looks like the divergence depends on how I write the electric field, but that does not make sense. Algebraically, my electric field and Griffiths' electric field are equal, and from a purely vector calculus point of view I cannot see anything wrong with my divergence either. Where is the my mistake?

 

[Rick16]

I have looked back at section 1.5 in the book, but it has not helped much so far. Apparently there are situations when the general expression for the divergence in spherical coordinates $$\nabla\cdot \vec v=\frac {1}{r^2}\frac {\partial}{\partial r}(r^2 v_r)+\frac {1}{r \sin \theta}\frac {\partial}{\partial \theta}(\sin \theta v_\theta)+\frac {1}{r \sin \theta}\frac {\partial v_\phi}{\partial \phi}$$ is not valid. How do I identify such situations? Is there a general strategy to follow?

 

[div_grad]

@Rick16 The way you calculated divergence is correct when you assume that $r \neq 0$, but this means that you have to exclude the point $r=0$ from the volume over which you integrate the density $\rho(r)$ when trying to get the total charge. Physically speaking, however, the singularity of the potential $V(r) \propto e^{-\lambda r}/r$ at $r=0$ means that there is a source (charge) there; furthermore, the exponential part $e^{-\lambda r}$ of the potential means that this charge at $r=0$ is screened by some other surrounding charges. Thus when you exclude the $r=0$ point from the volume integral, you obtain a non-zero total charge coming precisely from these surrounding screening charges.

If you were able to include the source charge at $r=0$ in the volume integral, then the total charge would be zero. However to do so, you cannot assume that $r \neq 0$ when calculating the divergence - and then you must use this formula involving a delta function.

Also note that you can use Gauss theorem to calculate the total charge directly from the electric field (we calculate over the surface of a large sphere with radius $R \rightarrow \infty$):
$$
Q = \varepsilon_0 \oint \boldsymbol{E}(R)\cdot\text{d}\boldsymbol{S} \rm{,}
$$
and since the electric field has the form $\boldsymbol{E} = f(r)\hat{\boldsymbol{r}}$ while the surface normal on the sphere is also directed along $\hat{\boldsymbol{r}}$ ($\text{d}\boldsymbol{S} = \hat{\boldsymbol{r}}\text{d}S = \hat{\boldsymbol{r}} R^2 \text{d}\Omega$, where $\text{d}\Omega$ is a solid angle element) , you have
$$
Q = \varepsilon_0 \oint f(r)\text{d}S = 4\pi\varepsilon_0 A e^{-\lambda R}(1+\lambda R) \rightarrow 0
$$
as $R \rightarrow \infty$, so you don't have to use $\rho(r)$ to get $Q$, in this case.

 

[div_grad]

@Rick16 Also, this term with a delta function can be obtained not from electric field but from the potential, since you have the Poisson equation $\nabla^2 V(r) = -\rho/\varepsilon_0$. Then you have
$$
\nabla^2 \frac{e^{-\lambda r}}{r} = e^{-\lambda r} \nabla^2\frac{1}{r} + \frac{1}{r}\nabla^2e^{-\lambda r} \rm{,}
$$
and then it is immediately seen that $\nabla^2 \frac{1}{r} = - 4\pi\delta(r)$ gives you this delta-function contribution to $\rho$, which you calculate from Poisson equation.

 

[Rick16]

Should I draw the conclusion that whenever I have a vector field in spherical coordinates with $r^2$ in the denominator, I cannot mechanically apply the above expression for the divergence, and that I should instead try to isolate $\frac{\hat r}{r^2}$, which then gives me an expression for the divergence including the delta function to take care of what happens at the origin? Would that be a general strategy?

 

[div_grad]

@Rick16 The general strategy would be to consider cases of $r \neq 0$ and $r = 0$ separately. The first case was already covered by your own calculation of $\nabla\cdot\mathbf{E}$ in Post #1, where you could differentiate everything without problems. In the second case, you investigate what happens at $r=0$ by surrounding this point with a small surface (for example, with a sphere with small radius $\delta \rightarrow 0$) and you calculate the flux of the field through this small surface (to do this, you use Gauss theorem to change the volume integral of divergence into surface integral of the normal component of the field).

For the particular case of the field in the form that you wrote it,
$$
\mathbf{E}(r) = A\left(\frac{\lambda e^{-\lambda r}}{r} + \frac{e^{-\lambda r}}{r^2}\right) \hat{\mathbf{r}} \rm{,}
$$
the case of $r\neq 0$ gives you your result $\rho(r) = -A\varepsilon_0 \frac{\lambda^2}{r} e^{-\lambda r}$. Now, to consider the $r=0$ case, surround the $r=0$ point with a small sphere of radius $\delta$ and calculate
$$
\varepsilon_0\int\nabla\cdot\mathbf{E}\,\,\text{d}^3r = \varepsilon_0\oint\mathbf{E}\cdot\text{d}\mathbf{S} = \varepsilon_0 A\left(\frac{\lambda e^{-\lambda \delta}}{\delta} + \frac{e^{-\lambda \delta}}{\delta^2}\right) 4\pi\delta^2 = 4\pi\varepsilon_0 A e^{-\lambda \delta}\left(\lambda \delta + 1\right) \rm{.}
$$
Then you let $\delta \rightarrow 0$ (small sphere!), which gives you the result
$$
\varepsilon_0\int\nabla\cdot\mathbf{E}\,\,\text{d}^3r \rightarrow 4\pi\varepsilon_0 A \rm{,}
$$
Since the left hand side above is formally (from Maxwell's equations) the volume integral $\int \rho \,\text{d}^3r$, you see that for it to be equal $4\pi\varepsilon_0 A$ you must set $\rho = 4\pi\varepsilon_0A\delta^3(\boldsymbol{r})$.

In this way you have investigated both the $r\neq 0$ and $r=0$ cases, so that for all values of $r$ the charge density is the sum of the results you obtained in both these cases:
$$
\rho(r) = 4\pi\varepsilon_0A\delta^3(\boldsymbol{r}) -A\varepsilon_0 \frac{\lambda^2}{r} e^{-\lambda r} = \varepsilon_0A\left(4\pi\delta^3(\boldsymbol{r}) - \frac{\lambda^2}{r} e^{-\lambda r}\right) \rm{.}
$$

 

[kuruman]

Should I draw the conclusion that whenever I have a vector field in spherical coordinates with $r^2$ in the denominator, I cannot mechanically apply the above expression for the divergence, and that I should instead try to isolate $\frac{\hat r}{r^2}$, which then gives me an expression for the divergence including the delta function to take care of what happens at the origin? Would that be a general strategy?

You should draw the conclusion that you need to be careful when using the expression for the divergence in spherical coordinates lest there be point charge singularities at the origin. In this particular case you have an electric field in the radial direction $$E_r=Ae^{-\lambda r}(1+\lambda r)\frac {1}{r^2}$$ and you need to take the radial partial derivative $$\frac{\partial}{\partial r}\left[r^2Ae^{-\lambda r}(1+\lambda r)\dfrac {1}{r^2}\right].$$ It is tempting but wrong to say at this point that $$\frac{\partial}{\partial r}\left[\cancel{r^2}Ae^{-\lambda r}(1+\lambda r)\dfrac {1}{\cancel{r^2}}\right]=\frac{\partial}{\partial r}\left[Ae^{-\lambda r}(1+\lambda r)\right]$$ because you will be throwing out the baby with the bath water.

Instead, consider one of the excellent suggestions provided by @div_grad. In addition to these, since you know the potential, you might also consider a multipole expansion which is easy because all there is to expand is the exponential. You have $$V(r)=\frac{A}{r}\left[1-\lambda r+\frac{1}{2}(\lambda r)^2-\dots\right]$$Clearly, the leading term is a monopole potential and signals "point charge $~q=4\pi \epsilon_0 A~$ at the origin."

 

[Rick16]

@Rick16 The general strategy would be to consider cases of $r \neq 0$ and $r = 0$ separately. The first case was already covered by your own calculation of $\nabla\cdot\mathbf{E}$ in Post #1, where you could differentiate everything without problems. In the second case, you investigate what happens at $r=0$ by surrounding this point with a small surface (for example, with a sphere with small radius $\delta \rightarrow 0$) and you calculate the flux of the field through this small surface (to do this, you use Gauss theorem to change the volume integral of divergence into surface integral of the normal component of the field).

For the particular case of the field in the form that you wrote it,
$$
\mathbf{E}(r) = A\left(\frac{\lambda e^{-\lambda r}}{r} + \frac{e^{-\lambda r}}{r^2}\right) \hat{\mathbf{r}} \rm{,}
$$
the case of $r\neq 0$ gives you your result $\rho(r) = -A\varepsilon_0 \frac{\lambda^2}{r} e^{-\lambda r}$. Now, to consider the $r=0$ case, surround the $r=0$ point with a small sphere of radius $\delta$ and calculate
$$
\varepsilon_0\int\nabla\cdot\mathbf{E}\,\,\text{d}^3r = \varepsilon_0\oint\mathbf{E}\cdot\text{d}\mathbf{S} = \varepsilon_0 A\left(\frac{\lambda e^{-\lambda \delta}}{\delta} + \frac{e^{-\lambda \delta}}{\delta^2}\right) 4\pi\delta^2 = 4\pi\varepsilon_0 A e^{-\lambda \delta}\left(\lambda \delta + 1\right) \rm{.}
$$
Then you let $\delta \rightarrow 0$ (small sphere!), which gives you the result
$$
\varepsilon_0\int\nabla\cdot\mathbf{E}\,\,\text{d}^3r \rightarrow 4\pi\varepsilon_0 A \rm{,}
$$
Since the left hand side above is formally (from Maxwell's equations) the volume integral $\int \rho \,\text{d}^3r$, you see that for it to be equal $4\pi\varepsilon_0 A$ you must set $\rho = 4\pi\varepsilon_0A\delta^3(\boldsymbol{r})$.

In this way you have investigated both the $r\neq 0$ and $r=0$ cases, so that for all values of $r$ the charge density is the sum of the results you obtained in both these cases:
$$
\rho(r) = 4\pi\varepsilon_0A\delta^3(\boldsymbol{r}) -A\varepsilon_0 \frac{\lambda^2}{r} e^{-\lambda r} = \varepsilon_0A\left(4\pi\delta^3(\boldsymbol{r}) - \frac{\lambda^2}{r} e^{-\lambda r}\right) \rm{.}
$$

This looks very convincing, but do you really go through all this everytime you are confronted with this kind of situation? Griffiths simply isolates the term $\frac{\hat r}{r^2}$ and then uses the relation $\nabla \cdot \frac{\hat r}{r^2}=4\pi \delta^3 (r)$. At first, I did not understand why he would do that, but I am now convinced of the necessity to keep the $\frac{1}{r^2}$ instead of throwing it out, which is what happens when you use the regular rule for the divergence, like kuruman explained:

It is tempting but wrong to say at this point that ∂∂r[r2Ae−λr(1+λr)1r2]=∂∂r[Ae−λr(1+λr)] because you will be throwing out the baby with the bath water.

It seems to me that Griffiths' method (isolating $\frac{\hat r}{r^2}$ and then using $\nabla \cdot \frac{\hat r}{r^2}=4\pi \delta^3 (r)$) should be generally applicable, when I would otherwise throw out the term $\frac{1}{r^2}$ by applying the regular divergence rule. Is this correct?

 

[kuruman]

@Rick16 Also, this term with a delta function can be obtained not from electric field but from the potential, since you have the Poisson equation $\nabla^2 V(r) = -\rho/\varepsilon_0$. Then you have
$$
\nabla^2 \frac{e^{-\lambda r}}{r} = e^{-\lambda r} \nabla^2\frac{1}{r} + \frac{1}{r}\nabla^2e^{-\lambda r} \rm{,}
$$
and then it is immediately seen that $\nabla^2 \frac{1}{r} = - 4\pi\delta(r)$ gives you this delta-function contribution to $\rho$, which you calculate from Poisson equation.

What happened to the cross term?
Isn't it true that $~\nabla^2(fg)=f\nabla^2g+2\mathbf{\nabla}f\cdot\mathbf{\nabla}g +g\nabla^2f~$?

 

[kuruman]

Is this correct?

Yes.

 

[vela]

This looks very convincing, but do you really go through all this everytime you are confronted with this kind of situation? Griffiths simply isolates the term r^r2 and then uses the relation ∇⋅r^r2=4πδ3(r). At first, I did not understand why he would do that, but I am now convinced of the necessity to keep the 1r2 instead of throwing it out, which is what happens when you use the regular rule for the divergence, like kuruman explained:

Griffiths went through the same derivation for $\nabla \cdot \frac{\hat r}{r^2}$ to show that there's a delta function at the origin. You can use this result without going through the derivation each time. Griffiths expressed the electric field in terms of $\nabla \cdot \frac{\hat r}{r^2}$ so he could use the previously derived result. If you don't write it that way, as you did, then you do need to show there's a delta function term, and @div_grad and @kuruman have showed you different ways to do this.

The lesson here is that whenever you have a division, you need to check for situations which result in a division by zero. In this problem, both the potential and electric field diverge as $r \to 0$, which suggests something wonky is going on at the origin which you need to investigate further. At the very least, it signals you can't naively differentiate either quantity at $r=0$.

 

[div_grad]

What happened to the cross term?
Isn't it true that $~\nabla^2(fg)=f\nabla^2g+2\mathbf{\nabla}f\cdot\mathbf{\nabla}g +g\nabla^2f~$?

Of course, I made a mistake. I was so eager to focus on the part with $\nabla^2 \frac{1}{r}$ that I forgot to write down the missing cross term, which you have now corrected.