How to Correctly Express Electric Charge Density for a Point Charge?

Click For Summary

Homework Help Overview

The discussion revolves around expressing the electric charge density for a point charge, specifically addressing the mathematical representation and dimensional analysis of the charge density function.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to formulate the charge density using a delta function but questions the dimensional consistency of their expression. Other participants suggest the correct form of the delta function and explore the implications of its dimensionality.

Discussion Status

Participants are actively engaging in clarifying the mathematical representation of the charge density. Some have provided insights into the dimensional analysis of the delta function, which has helped to address the original poster's confusion regarding units.

Contextual Notes

There is an emphasis on ensuring that the volume integral of the charge density equals the total charge, which raises questions about the dimensionality of the delta function in different dimensions.

Yeldar
Messages
6
Reaction score
0
(src: Intro to Electrodynamics, Griffith, Problem 1.46a)
Q: Write an expression for the electric charge density [itex]\rho (r)[/itex] of a point charge [itex]q[/itex] at [itex]r^'[/itex]. Make sure that the volume integral of [itex]\rho[/itex] equals [itex]q[/itex].

Now, Closest I can seem to come up with is:


[tex]\rho(r)=\frac{q}{4*Pi*R^2}\delta(r-r^')[/tex]

But, the problem I see with this, is that while yes, integrating this over any volume [itex]V[/itex] that enclosed the point charge will return q, but that q would have to have units of charge/unit_volume which just dosent make sense. Or am I missing something?

Any help would be appreciated.
 
Physics news on Phys.org
I think that delta function should be 3D: [itex]\delta^{(3)}(\vec{r}-\vec{r}')[/itex]. Note that the n-D delta function has dimensions of (length)-n.
 
Last edited:
Yeah, sorry missed that. Have the [itex]\delta^3[/itex] on my paper, just forgot to type it in.

I don't understand how n-D delta functions have a dimension of (length)-n, could you explain that perhaps?
 
Sure, let's look at the 1D case. Consider the following integral:

[tex]\int_{-\infty}^{\infty}\delta(x)dx=1[/tex]

The right side of that is 1. Not 1 meter or 1 Joule, just plain old 1. So if the units of dx are meters, then what must the units of the delta function be? Inverse meters.

Similar results hold for higher dimensional cases.
 
Okay, that makes sense.

Thanks for your help, this was driving me crazy, I couldn't figure out why units were not making sense.
 

Similar threads

Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
5
Views
2K
Replies
6
Views
2K
Replies
23
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
35
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
44
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K