I have two charged hemispheres (which are very close to each other, we consider them now as a sphere) with the charge density given as $\rho = \frac {Q}{\pi a^3 \frac {4}{3+n}} (\frac{r}{a})^n$, where $n \ge 0$. I want to find out how the charge density $\rho_{\infty}$ and the eletric field looks like for $n \to \infty$, where $r \lt a$ so $a$ is the radius of that sphere and $r$ is inside it.

The limit $n \to \infty$ for $\rho$ is obviously $0$, but I know I am going to use the Dirac delta function here. This function is zero everywhere outside the sphere, one everywhere within the sphere and infinity on the surface, where I am going to put the whole charge of that sphere for $\rho_{\infty}$. But I am not sure how to do it.

I have seen on physics stack some examples for the charged sphere and the Dirac delta function, but this example is specific because of that $\rho$ with $n$ power and because I am looking for $\rho_{\infty}$ and those examples didn't help me at all. I know I am not going to calculate that limit itself but I am going to use some integrals but I am actually not aware how could I do it. The only thing that came to my mind is:

$q_{\infty} = \int dq = \int_{-\infty}^{\infty} \delta(r-a)\rho dV$ where $q$ is a point charge and $V$ is a volume.

I don't know what to do with this. Many similar example were showing that $q=\rho$ or maybe $q = \rho \delta(a)$, I don't know.


1 Answer 1


So you have defined that $$\rho_n(r) = \begin{cases}\frac{(n+3)~Q}{4\pi~a^3}\left(\frac ra\right)^n&\text{if }r<a,\\~~~~0&\text{otherwise}.\end{cases}$$This does indeed have a fixed charge $Q$ on it. Indeed in the limit as $n$ goes to $\infty$ one has $(r/a)^n \to 0$ and this fixed charge will be concentrated in a thin shell near the surface of the ball. You are looking for a representation of this in terms of the Dirac $\delta$-function, which is not a function but something which is "function like" in that, though it cannot be graphed, it assigns a meaning to any integral that it appears in: this being that $$\int_a^bdx~\delta(x-x_0)~f(x) = \begin{cases}f(x_0) & \text{if }a < x_0 < b,\\ 0&\text{otherwise.}\end{cases}$$ But we have that the charge contained in the space between two spheres is, $$Q(r_0, r_1) = \int_{r_0}^{r_1}4\pi~r^2~dr~\rho_n(r).$$ If we want to express that this for $\rho_\infty$ is $\{Q\text{ if } r_0 < a < r_1 \text{ else } 0\}$ then clearly we need this expression to be $\int dr~\delta(r-a)~f(r)$ where $f(a) = Q.$ We have a tremendous amount of freedom in how we write this but perhaps the most simple would be $$\rho_N(r) = \frac{Q}{4\pi~a^2} ~\delta(r - a),$$ where we've used the fact that $r^2/a^2$ will be $1$ when evaluated at $a$. But you could also replace the $a^2$ with $r^2$ or $r~a$ or even something complicated like $\frac13 a^2 + \frac23 r~a$ and it would just be a symbolic reinterpretation of the same distribution. It's not entirely what one would call an exact science because the Dirac $\delta$-function is itself describing this thing that can be meaningfully integrated but not graphed, and there are many continuous functions which in some parameter's limit become the Dirac $\delta$-function.

The only real distinction worth making is perhaps the dimensionality of the parameter that the Dirac $\delta$-function accepts and hence that the integral is over; this is a 1D Dirac $\delta$-function $\delta^1$ which should be contrasted with the 3D Dirac $\delta$-function $\delta^3$ which we use for a point charge distribution. `

  • $\begingroup$ And If I have $\delta ^3$, how much different would it be? I actually have to use this $\delta ^3$, because it's a sphere. And so the charge density for the $n$ going to an infinity of the charged sphere is the regular charge density of the spherical shell? I am sorry for my not too much intelligent questions but I am drowning in it and thank you, your answer is very nice. :) $\endgroup$
    – user156350
    Sep 17, 2017 at 3:35
  • $\begingroup$ @Kris well you don't have a point charge, so you would have to think about a sum of point charges to use the $\delta^3$ expression as this charge density is something more like $\rho=\alpha~\int_{\mathcal S}d^2\vec r_0~\delta^3(\vec r - \vec r_0),$ overlaying a bunch of them on the sphere $\vec r_0\in\mathcal S.$ The connection here might be as simple as saying "well $d^2\vec r_0=a^2\sin\phi_0~d\phi_0~d\theta_0,$ act on this with $\int dr_0~\delta(r_0-a)$ to get an $\int d^3\vec r_0$ term, now you can evaluate that $\int d^3\vec r_0~\delta^3(\vec r-\vec r_0)~f(\vec r_0)$ to the above form." $\endgroup$
    – CR Drost
    Sep 17, 2017 at 17:43

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