Calculating Total Charge on Non-Uniformly Charged Sphere

[doktorwho]

Homework Statement

A sphere of radius $a$ is non-uniformly charged on its surface with a charge whose surface density is $ρ_s(φ)=ρ_{so}(cosφ)^2$ where $φ$ is the angle measures from the z axis, (0≤φ≤π) and $ρ_{s0}$ is a constant. Determine the expression for the total charge distributed on the sphere.

Homework Equations

$dQ=ρ_sdS$

The Attempt at a Solution

I know I am supposed to find the small surface element on which to integrate but the surface charge density is given by the angle and how am i supposed to make the surface element be in angle form. I tried thinking like this: In a circle the element $dL$ that is the small part of the circumference is $rdφ$ but don't know how to use that on the sphere..
The solution should be $Q=\frac{4π}{3}ρ_{s0}a^2$
The problem i have now is how to start. I have to find the surface element and i don't know how, can you help?

 

[TSny]

Welcome to PF!
Look at the last page of this link: http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf
But note that here they use the symbol $\theta$ for your $\varphi$.

 

[doktorwho]

Welcome to PF!
Look at the last page of this link: http://web.mit.edu/8.02t/www/materials/modules/ReviewB.pdf
But note that here they use the symbol $\theta$ for your $\varphi$.

Hi, its great to be here :D
I have solved the problem, i figured that the part of the sphere that is under the fixed angle can be integrated,
$dA=2rπdl$ where the circumference at some radius $r$ that is equal to $r=asinφ$ multiplied by the $dl$ element equaling to $adφ$ gives out the area and the integral becomes $Q=∫2πa^2ρ_{so}(cosφ)^2sinφ$ integrated on the interval $[0,π]$ but just out of curiosity how would i use the area you provided?
The surface element is $dA=a^2sinφdφdθ$ and the integral becomes $Q=∫ρ_{so}(cosφ)^2a^2sinφdφdθ$? There are two differentials now, how to use this?

 

[haruspex]

There are two differentials now, how to use this?

The integral with respect to θ is easy, so do that first.

 

[doktorwho]

The integral with respect to θ is easy, so do that first.

Well the limits are $[0, π]$ so it should be $π$ right? But then I am mising a factor of $2$ so it should be $2π$ somehow..

 

[haruspex]

the limits are [0,π]

In polar, to cover the sphere, one angle goes 0 to π and the other from 0 to 2π.

 

[doktorwho]

In polar, to cover the sphere, one angle goes 0 to π and the other from 0 to 2π.

So i integrate one angle from $[0, π]$ and the other $[0, 2π]$? How would i put the limits for the general expression?

 

[haruspex]

So i integrate one angle from $[0, π]$ and the other $[0, 2π]$? How would i put the limits for the general expression?

You are asking about the notation? $\int^{\pi}_{\phi=0}\int^{2\pi}_{\theta=0}$.

 

[doktorwho]

You are asking about the notation? $\int^{\pi}_{\phi=0}\int^{2\pi}_{\theta=0}$.

oh yeah it would be a double integral, thanks :D!