Volume Charge Density in Spherical and Cylindrical Coordinat

[RJLiberator]

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

Now, part b is pretty much the same thing.
$$ρ(r)=C \delta (r-R)$$
$$Q = \int ρda = \int C \delta (r-R) \pi r^2dr$$
$$C = \frac{Q}{\pi R^2}$$
$$ρ(r) = \frac{Q}{\pi R^2} \delta(r-R)$$Did I do everything right or am I missing something? I ask because it seems to 'easy' to be this identical.

 

[Charles Link]

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 $\theta_o$.

 

[RJLiberator]

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)
$$ρ(s) = \frac{q}{R} \delta (s-R) \delta (z)$$
From the integration of
$$Q = \int ρdv = \int C \delta (s-R) \delta (z) sdsdz$$
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)
$$ρ(r) = \frac{Q}{R^2} \delta(r-R) \delta (\theta - \theta_0)$$
From the integration of
$$Q = \int ρdv = \int C \delta(r-R) \delta( \theta - \theta_0) r^2sin(\theta)d \theta dr$$

In part b, I use dr instead of dΦ as there is dependence on r and theta, is that correct? (reference to your post)

 

[Charles Link]

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 $2 \pi$.

 

[RJLiberator]

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)
$$ρ(s) = \frac{q}{2 \pi R} \delta (s-R) \delta (z)$$
b)
$$ρ(r) = \frac{Q}{2 \pi R^2} \delta(r-R) \delta (\theta - \theta_0)$$

 

[Charles Link]

Very good ! Don't forget to put $\theta_o=\pi/2$ in your answer to (b).

 

[RJLiberator]

Thank you for the help! Thread is solved.