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Charge Density and Diract Delta Functions

  • #1

Homework Statement



a) A charge q1 = q is at r'1 = -Di, and a charge q2 = -3q is at r'2 = Di. Find an expression for the volume charge density p(r).

b) An infinitely long wire along the z-axis has a uniform linear charge density [tex]\lambda[/tex]. Find an expression for the volume charge density p(r) in cylindrical coordinates.


Homework Equations



[tex]\int[/tex][tex]\delta[/tex](x) dx = 1


The Attempt at a Solution



a) I'm not sure how to format my solution for the question, but I came up with this:
p(r) = [[tex]\delta[/tex](r - Di)(-3q) + [tex]\delta[/tex](r + Di)(q)]

b) I can't figure out how to do this with cylindrical coordinates. With Cartesian coordinates I came up with this:
p(x,y,z) = [tex]\delta[/tex](x)[tex]\delta[/tex](y)[tex]\lambda[/tex]

When I try with cylindrical coordinates, I realize that r must equal zero, so I know [tex]\delta[/tex](r) would be part of the expression; however, [tex]\theta[/tex] can be any value, as can z. That results in this:
p(r,[tex]\theta[/tex],z) = [tex]\delta[/tex](r)[tex]\lambda[/tex]
which has incorrect units for volume charge density.

Any help on this question is greatly appreciated!
 

Answers and Replies

  • #2
collinsmark
Homework Helper
Gold Member
2,880
1,181
Hello definerandom,

Welcome to Physics Forums!

If you want to get the units correct, perhaps consider using δ3(r) = δ(x)δ(y)δ(z) = δ(r)δ()δ(rsinθΦ) for points, and δ2(r) = δ(x)δ(y) = δ(r)δ() for lines (where the z component is not included in r in the second case).

[Edit: Modified the cylindrical and polar representations. Anyway, the point being is that you can square or cube the Dirac delta function as appropriate, noting that each Dirac delta function has units of 1/length (such as 1/meters, for example.)

[Another edit: So in other words, when dealing with points, you should end up with δ3(), which has units of length-3. When working with lines, δ2() which has units of length-2.]
 
Last edited:

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