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I just got introduced to the Dirac delta function and one of the questions was to express volume charge density $\rho({\bf r})$ of a point charge $q$ at origin. I saw that the answer is related to Dirac delta function as: $$\rho({\bf r}) = q \delta^3({\bf r})$$ where $\delta^3({\bf r})$ is the 3-dimensional Dirac Delta function. Why is it so?

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  • $\begingroup$ Why is it cubed? That doesn't really make any sense to cube a delta function. $\endgroup$ – Goldname Jan 28 '17 at 2:50
  • $\begingroup$ @Goldname : Not a cube ,that represents three dimensional delta function :) $\endgroup$ – Lapmid Jan 28 '17 at 3:02
  • $\begingroup$ @Theasgardian Wow mind blown $\endgroup$ – Goldname Jan 28 '17 at 3:14
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    $\begingroup$ Every time you see Dirac delta functions, you should think of them as being integrated. Otherwise, it says that the charge density at that point is some sort of infinity. Integrating the Dirac delta will give you the total charge in a certain volume. $\endgroup$ – Ben S Jan 28 '17 at 3:19
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A point charge is confined to a single point in space. Let's call it as $q\left(\vec{r}\right)$. This means that the charge has a magnitude only at $\vec{r}$, and at all other points it is zero.

The charge density for a point charge is given by charge per unit volume. Since the there is no charge except at the position $\vec{r}$, the charge density vanishes at all points except $\vec{r}$. Now, at $\vec{r}$, the volume gives off to the limit $V\rightarrow0$. So the charge density blows up to infinity at that point. It's the case for all point sources, not just restricted to point charges: for example, the mass density of a point mass blows up at the origin.

Does this work for all point charges? Yes. But, there is an exception when the magnitude of the point charge is zero. However if that's the case, why should one be concerned about all this? So, a point charge has a non-zero magnitude at the point it occupies. So, it works all time.

This particular property of the charge density of a point charge is exactly identical to the definition of the Dirac-delta function, which, for the point $\vec{r}$ can be defined as
$$ \delta^3\left(\vec{r}\right) = \begin{cases} \infty, & \text{at the point $\vec{r}$} \\[2ex] 0, & \text{at all other points} \end{cases} $$

So, it seems quite comfortable that we could use this function to represent the volume charge density of a point charge. It is to be noted that at all other points except at $\vec{r}$, the charge is zero (which contributes to the second case of the Dirac-delta function defined above, and at the point $\vec{r}$, the first case of Dirac- delta function is satisfied by the inverse volume.

However, your question is concerned by the definition of the charge density of a point charge at the "origin" (here $\vec{r}$). So, at the origin, the charge is non-zero, but the inverse volume blows up. So, we substitute the Dirac-delta function in the place of inverse volume as

$$\rho=q\delta^3\left(\vec{r}\right)$$

This definition is however valid at other points than origin also, since at all other points the delta function vanishes and so does the charge density. But for a point charge, the result is trivial that $q=0$ at all other points.

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    $\begingroup$ Such a wonderful insight. I feel much comfortable with the idea of Dirac-Delta function now. Thanks a lot! $\endgroup$ – Ufomammut Jan 28 '17 at 5:58
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If you integrate $\rho$ over the volume, the $\delta$ will be zero everywhere except where its argument is $0$. Since the area under this function is normalized to $1$, the result of the integration will be that the charge will be $0$ everywhere except at $r=0$ where it will be $q$.

This follows because you want $\rho(r)$ to be $0$ everywhere except at $r=0$ (hence the $\delta(r)$), and you want $\int dV \rho(r)=q$ since this is the only charge in your system, hence the $q\delta(r)$.

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  • $\begingroup$ I get that part. My only confusion is let's say I don't even know the expression for rho(r). How do I approach the problem? What factors should I look for? $\endgroup$ – Ufomammut Jan 28 '17 at 2:33
  • $\begingroup$ you have the expression for $\rho: q\delta(r)$. $\endgroup$ – ZeroTheHero Jan 28 '17 at 2:36
  • $\begingroup$ @Avery : I just added another bit. $\endgroup$ – ZeroTheHero Jan 28 '17 at 2:40
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It has been explained in other answers nicely. I want to add a little bit. Since you said that, Dirac Delta has just been introduced, you must have come around the fact that Dirac Delta is also a distribution.

So always remember that thing. In this case, Dirac Delta used with charge density shows how charge is distributed. Thus, for a point charge, it is zero everywhere except at the source position, where it is maximum(in all three dimensions).

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