Find the total electric charge in a spherical shell

[Physics Dad]

Homework Statement

Find the total electric charge in a spherical shell between radii a and 3a when the charge density is:

ρ(r)=D(4a-r)

Where D is a constant and r is the modulus of the position vector r measured from the centre of the sphere​

Homework Equations

Q=ρV

Volume of a sphere = (4/3)πr³

The Attempt at a Solution

My initial thinking was that I needed to get involved with the different forms of Gauss' Law, however the more I think about it the less I understand.

With the statement being a shell, should I consider two Gaussian surfaces at the various r values and sum the two charge values? Or should I do as I have done here and assume volumes as I have a charge density?

My attempt is:

When r=a

ρ(r) = D(4a-a) = 3Da

Q₁ = ρV = (3Da)(4/3πa³)
Q₁ = 4πDa⁴

When r = 3a

p(r) = D(4a - 3a) = Da

Q₂ = ρV = (Da)(4/3π(3a)³)
Q₂ = 12πDa⁴

Total charge = Q₂-Q₁ = 8πDa⁴

Now common sense is screaming at me saying this is wrong, but I am unsure where I should be going if this is the case.

Would love your feedback.

Thanks in advance.

 

[mfb]

Q=ρV

That formula only works for constant charge density.
The question asks about the total charge, not about electric fields, Gauss' law is not relevant here.

You'll need an integral.

 

[Physics Dad]

I thought as much,

So am I looking at:

Q = ∫ ρ(r) dV

and if so, is it:

Q = ^3r∫_r ρ(r) 4/3πr³ dr ?

Many thanks

 

[haruspex]

I thought as much,

So am I looking at:

Q = ∫ ρ(r) dV

and if so, is it:

Q = ^3r∫_r ρ(r) 4/3πr³ dr ?

Many thanks

Not quite.
V(r)=4/3πr³, so what is dV?

 

[Physics Dad]

ahh, right, so spherical polar?

Q = ^π∫₀ ^2π∫₀ ^3r∫_r ρ(r) r²sinφ dr dφ dθ ?

 

[haruspex]

ahh, right, so spherical polar?

Q = ^π∫₀ ^2π∫₀ ^3r∫_r ρ(r) r²sinφ dr dφ dθ ?

No need for that. Just answer my simple calculus question in post #4.

 

[Physics Dad]

OK, so dV is the change in volume which would be 4/3π(3a)³-4/3πa³

so dV=4/3π(2a)³?

so am I right in thinking this basically becomes a constant and can come out of the integral along with the D in ρ(r)?

If so, I get...

Q=4/3π(2a)³D ^3a∫_a 4a-r dr

Am I right to here or have I done something stupid?

Again, thank you!

 

[haruspex]

OK, so dV is the change in volume which would be 4/3π(3a)3-4/3πa3

No, I mean dV as in a derivative. $\int \rho(r).dV=\int \rho(r)\frac{dV}{dr}.dr$.
What is dV/dr?

 

[Physics Dad]

you mean 4πr²?

 

[haruspex]

you mean 4πr²?

Yes!

 

[Physics Dad]

OK,

so its...

Q = ^3a∫_a D(4a-r)⋅4πr² dr

which is the same as...

Q = 4πD ^3a∫_a 4ar² -r³ dr

Is this right so far?

Really sorry, I am sure you're pulling your hair out at me here!

 

[Physics Dad]

Its ok, I know I am wrong there, integration by parts!

 

[Physics Dad]

OK,

So I have ended up with an answer of...

Q=(176/3)πDa⁴

Would really appreciate confirmation either way.

Thank you so much for your help, I really appreciate it.

 

[haruspex]

Q=(176/3)πDa⁴

That's what I get.

 

[Physics Dad]

Thank you. You have the patience of a Saint!

 

[haruspex]

Thank you. You have the patience of a Saint!

Did you understand what I did in post #8?

 

[Physics Dad]

Yes, I did. I was just thinking about it in completely the wrong way.