# How to determine charge density using Dirac deltas in advance? — not after the fact

Context

I have already asked one question regarding charge densities, Diracs, and Heavisides . At the time of writing, that question remains open. More importantly, I still remain unclear regarding how to write charge densities with Diracs, and Heavisides. Here is an example to show that I lack clarity.

This question here revolves around how to define a surface charge. There is a large set up to the problem. Once set up, then comes the discussion of surface charge. I want to solve for the electric potential $$V$$ inside and outside a conductive ball of radius $$R$$ with a total charge $$Q$$. The problem can be solved with Gauss' Law, and it can be solved with spherical harmonics. I will do both.

Gauss' Law

Here, I use Gauss' Law, to find the electric field; and then I use integration and continuity of $$V$$ at the boundary to solve for $$V$$. I the potential at infinity as the zero referene. \begin{align*} \mathbf{E} {\left(\mathbf{r}\right)} &= \begin{cases} \mathbf{0},~&\text{for} ~r< R \\\\ \frac{Q}{4\,\pi\,\epsilon_o \,r^2 }\,\mathbf{\hat{r}} ,~&\text{for} ~r> R. \end{cases} \end{align*} \begin{align*} V {\left(\mathbf{r}\right)} &= \begin{cases} \frac{Q}{4\,\pi\,\epsilon_o \,R } ,~&\text{for} ~r\leq R \\\\ \frac{Q}{4\,\pi\,\epsilon_o \,r } ,~&\text{for} ~r> R. \end{cases} \end{align*}

Expansion in Spherical Harmonics

In a nut shell, since this problem has azimuthal symmetry, the solution can be written  in terms of Legendre polynomials of degree $$\ell$$, $$P_\ell$$, as $$V(r,\theta,\phi) = \begin{cases} \sum_{\ell=0}^\infty A_\ell\, r^\ell\,P_\ell(\cos\theta), &(r\leq R) \\ \sum_{\ell=0}^\infty \frac{A_\ell\,R^{2\,\ell+1}}{r^{\ell+1}} \,P_\ell(\cos\theta), &(r\geq R) ; \end{cases}$$ where $$A_\ell = \frac{1}{2\,\varepsilon_o\,R^{\ell-1} }\,\int_0^\pi \sigma(\theta) P_\ell(\cos\theta)\,\sin\theta\,d\theta .$$

Question

(1) What is $$\sigma(\theta)$$

(2) What am I misunderstanding about how to use Dirac delta distributions and Heaviside step function as I attempt to write charge densities in terms of the same?

Answer to Question 1

One the one hand, I argue, but not with self confidence and only because I know the answer in advance, that surface charge density is invariant with respect to $$\theta$$ (i.e., $$\sigma(\theta) = \sigma_o$$). In such case, according to the orthogonality of the Legendre polynomials  \begin{align} A_\ell &= \frac{\sigma_o}{2\,\varepsilon_o\,R^{\ell-1} }\,\int_0^\pi P_\ell(\cos\theta)\,\sin\theta\,d\theta . \\ &= \begin{cases} 0 & (\ell\neq 0) \\ \frac{\sigma_o\,R }{ \varepsilon_o }. \end{cases} \end{align} Thus, $$V(r,\theta,\phi) = \begin{cases} \frac{\sigma_o\,R }{ \varepsilon_o } , &(r\leq R) \\ \frac{\sigma_o\,R }{ \varepsilon_o }\,\frac{ R }{r }, &(r\geq R) ; \end{cases}$$ It appears that for the answer here to be self-consistent with the answer from Gauss' law, that $$\sigma(\theta) = \frac{Q}{4\,\pi\,R^2}.\tag{1}$$ This seems like a sensible description of the surface charge. Ulitmately, $$V(r,\theta,\phi) = \begin{cases} \frac{ Q }{ 4\,\pi\,R \,\varepsilon_o } , &(r\leq R) \\ \frac{ Q }{ 4\,\pi \,\varepsilon_o\,r }, &(r\geq R) . \end{cases}$$ Well, that appears to have went well.

One the other hand, I argue that surface charge density varies with respect to $$\theta$$. In such case, according to the orthogonality of the

I write that \begin{align} Q &= \int_{\mathbb{R}^3} \rho(r,\theta,\phi)\,d\tau \\ &= \int_{\phi=0}^{2\,\pi}\,\int_{\theta=0}^{\pi}\,\int_{r=0}^\infty \rho(r,\theta,\phi)\,r^2\,\sin\theta\,dr\,d\theta\,d\phi. \end{align} Since the charge only exists at $$r=R$$, I write $$\rho(r,\theta,\phi) = \sigma(r,\theta,\phi)\,\delta(r-R)$$, then \begin{align} Q &= \int_{\phi=0}^{2\,\pi}\,\int_{\theta=0}^{\pi}\,\int_{r=0}^\infty \sigma( \theta,\phi)\,\delta(r-R)\,r^2\,\sin\theta\,dr\,d\theta\,d\phi. \\ &= \int_{\phi=0}^{2\,\pi}\,\int_{\theta=0}^{\pi} \sigma(\theta,\phi) \, R^2\,\sin\theta\,dr\,d\theta\,d\phi . \end{align} I argue, but not self-convincingly, that since the differential area element, $$da$$ is $$da =R^2\,\sin\theta\,d\theta\,d\phi$$ and since the surface charge density should be constant with respect to $$da$$, that $$\sigma( \theta,\phi) =\sigma_o$$. Then, \begin{align} Q &= \int_{\phi=0}^{2\,\pi}\,\int_{\theta=0}^{\pi} \sigma_o \, R^2\,\sin\theta\,dr\,d\theta\,d\phi . \\ &= 4\,\pi \sigma_o \, R^2. \end{align} Again, in oder to be consistent, it seems that $$\sigma = \frac{Q}{4\,\pi\,R^2}. \tag{2}$$ In such case, I will get the same answer as before for $$V(r,\theta,\phi)$$.

My true question

Is there some way to more formal method to determine how to use Dirac delta distributions and Heavside step functions? Both ways that I performed here are round about. In Equation 1 and Equation 2, I determine the charge density after the fact. I do not like that, as I believe that I will fail to properly apply Dirac delta functions and Heaviside step functions on more complicated surfaces---when I can't use an after-the-fact argument.

Bibliography

 Griffiths, "Introdution to Electrodynamics" 2nd Edition p. 142-143.

 Wikipedia contributors. "Legendre polynomials." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 21 Feb. 2021. Web. 8 Mar. 2021.

• I don't understand what your question is. The uniform surface charge density is total charge/area, that's it. In any case, $da$ should not include $dr$. – fqq Mar 8 at 16:17
• As a side note, the phrase is ex post facto, from the Latin phrase meaning "retroactively." – J. Murray Mar 8 at 16:33

## 1 Answer

One the one hand, I argue, but not with self confidence and only because I know the answer in advance, that surface charge density is invariant with respect to $$\theta$$.

Your solution using Gauss' law implicitly assumes spherical symmetry. If $$\sigma(\theta)$$ varies with $$\theta$$, then $$\mathbf E \neq \frac{Q}{4\pi \epsilon_0 r^2} \hat r$$.

In any case, if your surface charge density on the surface of the sphere is $$\sigma(\theta)$$, then the spatial charge distribution (in 3D) is given by $$\rho(\mathbf r) = \sigma(\theta) \delta(r-R) = \frac{Q}{4\pi R^2} \delta (r-R)$$. One can see immediately that this works, because

$$Q_{inside} = \int _0^{r_0} r^2 \mathrm dr \int_0^{2\pi} \mathrm d\phi \int_{-1}^1 \mathrm d(\cos(\theta)) \ \left[\frac{Q}{4\pi R^2} \delta(r-R)\right] = \begin{cases}0 & r_0 R\end{cases}$$

On a more complicated surface, perhaps defined by $$R=R(\theta, \phi)$$, then if the surface charge distribution is given by $$\sigma(\theta,\phi)$$, the volume charge distribution will be given by $$\rho(r,\theta,\phi) = \sigma(\theta,\phi) \delta\big(r - R(\theta,\phi)\big)$$.

• Regarding Gauss law, you are quite right. I overlooked that. – Michael Levy Mar 8 at 16:56
• Ok, so how about demisphere of charge? Would you you utilize step fuctions, how would you right $\rho(r,\theta,\phi)$ then? – Michael Levy Mar 8 at 16:58
• @MichaelLevy What do you mean by a demisphere of charge? If e.g. the southern hemisphere has no charge on it, then $\sigma(\theta)=0$ for $\theta > \pi/2$. – J. Murray Mar 8 at 17:06
• How about uniform charge distribution on a demisphere given by $\rho(r,\theta,\phi) = \rho_o \,\left[u\left(\theta,\frac{\pi}{4}\right)-u\left(\frac{3\,\pi}{4},\theta,\right)\right] \,\left[u\left(\phi,\frac{\pi}{8}\right)-u\left(\frac{3\,\pi}{8},\phi,\right)\right]\,\delta(r,R(\theta,\phi))$. Is this a uniform charge density on a demisphere? – Michael Levy Mar 8 at 17:14
• @MichaelLevy If you wish, the uniform charge distribution on the northern hemisphere (the word demisphere is not in common usage) can be written as above by letting $\sigma(\theta) =\sigma_0 u(\frac{\pi}{2}-\theta)$, where $u$ is the Heaviside step function. Alternatively, you could just define it piecewisely as $\sigma_0$ for $\theta<\pi/2$ and $0$ for $\theta>\pi/2$; these are two different ways to write precisely the same thing. – J. Murray Mar 8 at 17:22