Charge density in terms of (r,θ) but need it in terms of the vector r'

[darkpsi]

Homework Statement

A sphere of radius R, centered at the origin, carries charge density
ρ(r,θ) = (kR/r²)(R - 2r)sinθ,
where k is a constant, and r, θ are the usual spherical coordinates.
Find the approximate potential for points on the z axis, far from the sphere.

Homework Equations

potential of the dipole
V(r,θ) = 1/4πε_o * 1/r² ∫ r'cosθ' ρ(r') dτ'

The Attempt at a Solution

I just want to know why I'm giving the charge density in terms of r and θ and yet I need it in terms of r'. I know that the vector r' is related to θ' by r' = rcosθ'
So I tried to say that since θ is the inclination from the xy plane and θ' is the angle between r and r' so that θ'+θ = π/2. Then that would give me
sinθ = [(π/2))-θ'] = cosθ'
Since the direction of r' is cosθ' I thought maybe I could replace r with r'?
so maybe
ρ(r',θ') = (kR/r'²)(R - 2r')cosθ' so
ρ(r') = (kR/r'²)(R - 2r') ?
PLEASE help I have a test on this tomorrow..

 

[jdwood983]

Homework Statement

A sphere of radius R, centered at the origin, carries charge density
ρ(r,θ) = (kR/r²)(R - 2r)sinθ,
where k is a constant, and r, θ are the usual spherical coordinates.
Find the approximate potential for points on the z axis, far from the sphere.

Homework Equations

potential of the dipole
V(r,θ) = 1/4πε_o * 1/r² ∫ r'cosθ' ρ(r') dτ'

The Attempt at a Solution

I just want to know why I'm giving the charge density in terms of r and θ and yet I need it in terms of r'. I know that the vector r' is related to θ' by r' = rcosθ'
So I tried to say that since θ is the inclination from the xy plane and θ' is the angle between r and r' so that θ'+θ = π/2. Then that would give me
sinθ = [(π/2))-θ'] = cosθ'
Since the direction of r' is cosθ' I thought maybe I could replace r with r'?
so maybe
ρ(r',θ') = (kR/r'²)(R - 2r')cosθ' so
ρ(r') = (kR/r'²)(R - 2r') ?
PLEASE help I have a test on this tomorrow..

Looks to me that you are confusing yourself on a simple matter. When inside the integral, it is often convenient to use primed-coordinates to separate from the original matter. If you had

<br /> V(r,\theta)=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^2}\int r\cos[\theta]\rho(r)d\tau<br />

it might confuse you because your limits of integration would also be in terms of r and \theta. The prime markers just show that, inside the integrand, r&#039; is slightly different from r, but does become r after integration; this way you have

<br /> V(r,\theta)=\frac{1}{4\pi\varepsilon_0}\,\frac{1}{r^2}\int r&#039;\cos[\theta&#039;]\rho(r&#039;)d\tau&#039;<br />

to tell the difference between what needs to be integrated and what is already taken care of through integration.

 

[darkpsi]

so in ρ(r,θ) = (kR/r2)(R - 2r)sinθ, r and θ are bascially r' and θ'?
when I integrate it that way I'm getting
∫ r'cosθ' kR(R-2r')sin²θ' dr'dθ'dφ'
then integrating ∫sin²θ' cosθ' dθ' I get
1/3sin³θ' (0 -> pi) which gives me zero.
And the answer is supposed to be
V(z) = 1/4πε_o kπ²R⁵/(24z²)

 

[gabbagabbahey]

And the answer is supposed to be
V(z) = 1/4πε_o kπ²R⁵/(24z²)

What makes you think that?...I also get that the monopole and dipole terms both vanish, and so the dominant term will be the quadrapole term...

 

[darkpsi]

But then the answer I'm supposed to be aiming for is wrong? At least I know maybe my qudrapole term wasn't wrong then, and maybe it's right?
3πkR5/64εoz3 ?
I got the anwser that it is suppose to be of this website http://einstein1.byu.edu/~masong/emsite/S2Q50/S2Q50.html

 

[gabbagabbahey]

That's a pretty sketchy website to be looking for homework solutions...the answer they give doesn't even have the correct units, so that should be a dead giveaway that something is wrong.

Your quadrapole term is a little off...why not show me your calculation?

 

[darkpsi]

Here goes:

V(r,θ) = 1/4πε_o * 1/r³ ∫∫∫₀^R (3/2 cos²θ' - 1/2) kRsin²θ' (Rr'² - 2r'³) dr'dθ'dφ'

= 1/4πε_o * 1/r³ ∫∫₀^π (3/2 cos²θ' - 1/2) kRsin²θ' (-1/6 kR⁴) dθ'dφ'

| Using 3/2 cos²θ'(cos²θ' - 1)
| = 3/2 (cos⁴θ' - cos²θ')
| = 3/2 [(1/2 + 1/2cos2θ')² - cos²θ']

= 1/4πε_o * 1/r³ ∫∫₀^π [3/2 [(1/2 + 1/2cos2θ')² - cos²θ'] sin²θ' (-1/6 kR⁴) dθ'dφ'

= 1/4πε_o * 1/r³ ∫∫₀^π 3/2[(1/4 + 1/2cos2θ' + 1/4cos²2θ')- cos²θ'] -1/2sin²θ' (-1/6 kR⁴) dθ'dφ'

= 1/4πε_o * 1/r³ (-1/6 kR⁴) ∫₀^2π( 3/2[(θ'/4 + 1/4sin2θ' + θ'/8 + 1/16sin2θ')] - 1/2(θ'/2 - 1/4sin2θ'|₀^π dφ'

| Since all the sine terms are zero I got

= 1/4πε_o * 1/r³ (-1/6 kR⁴) ∫₀^2π( 3/2[(θ'/4 + θ'/8)] - 1/2(θ'/2)|₀^π dφ'

= 1/4πε_o * 1/r³ (-1/6 kR⁴) ∫₀^2π 3/2(-3/8π) dφ'

= 1/4πε_o * 1/r³ (-1/6 kR⁴) 3/2(-3/8π) * 2π

= 3πkR⁵ / 64ε_or³ |_r=z

= 3πkR⁵ / 64ε_oz³



I think there may be a mistake in my use of the half-angle identity?

 

[gabbagabbahey]

| Using 3/2 cos²θ'(cos²θ' - 1)
| = 3/2 (cos⁴θ' - cos²θ')
| = 3/2 [(1/2 + 1/2cos2θ')² - cos²θ']

Why are you using this?

\left(\frac{3}{2}\cos^2\theta&#039;-\frac{1}{2}\right)=\frac{1}{2}(3\cos^2\theta&#039;-1)=\frac{1}{2}[3(1-\sin^2\theta&#039;)-1]=1-\frac{3}{2}\sin^2\theta&#039;\neq\frac{3}{2}\cos^2\theta&#039;(\cos^2\theta&#039;-1)

 

[darkpsi]

Why are you using this?

\left(\frac{3}{2}\cos^2\theta&#039;-\frac{1}{2}\right)=\frac{1}{2}(3\cos^2\theta&#039;-1)=\frac{1}{2}[3(1-\sin^2\theta&#039;)-1]=1-\frac{3}{2}\sin^2\theta&#039;\neq\frac{3}{2}\cos^2\theta&#039;(\cos^2\theta&#039;-1)

Because I was distributing the sin² to just that term and then trying to solve that integral. But I see that you're way is much easier. Thanks!