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Context

I have already asked one question regarding charge densities, Diracs, and Heavisides [0]. 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 [1] 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 [2] \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

[0] Dirac delta, Heaviside step, and volume charge density

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

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

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    $\begingroup$ 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 $dr$. $\endgroup$ – fqq Mar 8 at 16:17
  • $\begingroup$ As a side note, the phrase is ex post facto, from the Latin phrase meaning "retroactively." $\endgroup$ – J. Murray Mar 8 at 16:33
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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

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

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 $\rho(r,\theta,\phi) = \sigma(\theta,\phi) \delta\big(r - R(\theta,\phi)\big)$.

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  • $\begingroup$ Regarding Gauss law, you are quite right. I overlooked that. $\endgroup$ – Michael Levy Mar 8 at 16:56
  • $\begingroup$ Ok, so how about demisphere of charge? Would you you utilize step fuctions, how would you right $\rho(r,\theta,\phi)$ then? $\endgroup$ – Michael Levy Mar 8 at 16:58
  • $\begingroup$ @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 $\theta > \pi/2$. $\endgroup$ – J. Murray Mar 8 at 17:06
  • $\begingroup$ 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? $\endgroup$ – Michael Levy Mar 8 at 17:14
  • $\begingroup$ @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. $\endgroup$ – J. Murray Mar 8 at 17:22

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