Volume Charge Density in Spherical and Cylindrical Coordinat

In summary, we learned how to find the volume charge density for a ring of radius R with a total charge of Q uniformly distributed on the ring. In cylindrical coordinates, the volume charge density is expressed as ρ(s) = Q/2πR * δ(s-R) * δ(z). In spherical coordinates, the volume charge density is expressed as ρ(r) = Q/2πR^2 * δ(r-R) * δ(θ-π/2). These calculations were done by integrating over the appropriate volume element and taking into account the dependencies on the different variables.
  • #1
RJLiberator
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Homework Statement


Consider a ring of radius R placed on the xy-plane with its center at the origin. A total charge of Q is uniformly distributed on the ring.
a) Express the volume charge density of this configuration ρ(s,Φ,z) in cylindrical coordinates.
b) Express the volume charge density of this configuration ρ(r,Θ,Φ) in spherical coordinates.

Homework Equations

The Attempt at a Solution


I am going to show you the work that I did.

a) [tex]ρ(s) = C\delta (s-R)[/tex]
And so, we need to find the constant C.
[tex]Q = \int ρda = \int C \delta (s-R) \pi R^2 dR[/tex]
Thus ##C = \frac{Q}{\pi R^2}## and we see
[tex]ρ(s) = \frac{Q}{\pi R^2}\delta(s-R)[/tex]

Now, part b is pretty much the same thing.
[tex] ρ(r)=C \delta (r-R)[/tex]
[tex] Q = \int ρda = \int C \delta (r-R) \pi r^2dr[/tex]
[tex]C = \frac{Q}{\pi R^2}[/tex]
[tex] ρ(r) = \frac{Q}{\pi R^2} \delta(r-R)[/tex]Did I do everything right or am I missing something? I ask because it seems to 'easy' to be this identical.
 
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  • #2
You need to do these integrals over ## dv ##, and they both need a couple of corrections. ## \\ ## a) For cylindrical coordinates, ## dv=s \,ds \, d\Phi \, dz ##. There is no ## \Phi ## dependence, but you will need a ## \delta(z) ## function in addition to the ## \delta(s-R) ##. ## \\ ## b) For spherical coordinates, ## dv=r^2 \, sin(\theta) \, d \theta \, d\phi \, dr ##. A hint in part (b) is you will need a ## \delta(\theta-\theta_o) ## type delta function for some ## \theta_o ##.
 
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  • #3
Charles, thanks for the help.

I see that it needs to be in volume due to the volume charge density. I was messed up as it is a ring, so I thought it'd just be area.

With your help/hints, I think I have the following:
a)
[tex] ρ(s) = \frac{q}{R} \delta (s-R) \delta (z)[/tex]
From the integration of
[tex]Q = \int ρdv = \int C \delta (s-R) \delta (z) sdsdz [/tex]
Which evaluates at s = R, and z =0, thus leaving CR = Q and C = Q/R which is where we get the final ρ(s).

b)
[tex] ρ(r) = \frac{Q}{R^2} \delta(r-R) \delta (\theta - \theta_0) [/tex]
From the integration of
[tex] Q = \int ρdv = \int C \delta(r-R) \delta( \theta - \theta_0) r^2sin(\theta)d \theta dr[/tex]

In part b, I use dr instead of dΦ as there is dependence on r and theta, is that correct? (reference to your post)
 
  • #4
For (a) you left off the ## d \Phi ## in the ## dv ## with its integration. You also left off the ## d \phi ## in the ## dv ## in the integration of part (b). For part(b), you should be able to give the value of ## \theta_o ## based on the spherical coordinates that will make the ring in the x-y plane. The ## d \theta ## integration will subsequently get you a ## sin(\theta_o) ## term, but since you know ## \theta_o ##, you also know ## sin(\theta_o) ##. ## \\ ## And a hint for the ## d \phi ## integrations is that in both (a) and (b) it goes from ## 0 ## to ## 2 \pi ##.
 
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  • #5
I see.
I had thought that due to the dirac delta function and the "lack of dependecny" on dΦ that I wouldn't need them. But it is clear that there is a dependency on this volume charge.
The thing I needed to add to both integrals was dΦ and this goes from 0 to 2pi as it is a ring, this results in the 2pi factor being included in my answers now.

As for the ##sin( \theta_0 )## term, we see that the dirac delta function for that term is equal to 0 when ##( \theta - \pi/2)## is equal to 0. So ##\theta = \pi /2##. Now, ##sin( \pi /2) = 1## thus it is a non-factor in the final volume distribution.

a)
[tex]
ρ(s) = \frac{q}{2 \pi R} \delta (s-R) \delta (z)[/tex]
b)
[tex]
ρ(r) = \frac{Q}{2 \pi R^2} \delta(r-R) \delta (\theta - \theta_0) [/tex]
 
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  • #6
Very good ! Don't forget to put ## \theta_o=\pi/2 ## in your answer to (b).
 
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  • #7
Thank you for the help! Thread is solved.
 
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FAQ: Volume Charge Density in Spherical and Cylindrical Coordinat

1. What is volume charge density in spherical and cylindrical coordinates?

Volume charge density in spherical and cylindrical coordinates is a measure of the electric charge present in a given volume. It is represented by the Greek letter rho (ρ) and is expressed in units of coulombs per cubic meter (C/m3).

2. How is volume charge density calculated in spherical coordinates?

In spherical coordinates, volume charge density is calculated by dividing the total charge in a spherical volume by the volume itself. This can be represented mathematically as ρ = Q/V, where ρ is the volume charge density, Q is the total charge, and V is the volume in question.

3. What is the difference between volume charge density in spherical and cylindrical coordinates?

The main difference between volume charge density in spherical and cylindrical coordinates is the shape of the volume being considered. In spherical coordinates, the volume is a sphere, while in cylindrical coordinates, it is a cylinder. This affects the way the charge is distributed and therefore the calculation of volume charge density.

4. How does volume charge density affect electric fields in spherical and cylindrical coordinates?

Volume charge density is directly related to the strength of the electric field in a given region. In both spherical and cylindrical coordinates, a higher volume charge density results in a stronger electric field. This is because the electric field is determined by the amount of charge present in a given volume.

5. Can volume charge density be negative?

Yes, volume charge density can be negative if the total charge in a given volume is negative. This can occur when there is an excess of negative charges in a region, resulting in a negative volume charge density. However, in most cases, volume charge density is positive, as most objects have an overall positive charge.

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