# Find the total electric charge in a spherical shell

## 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.

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mfb
Mentor
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.

I thought as much,

So am I looking at:

Q = ∫ ρ(r) dV

and if so, is it:

Q = 3rr ρ(r) 4/3πr3 dr ?

Many thanks

haruspex
Homework Helper
Gold Member
I thought as much,

So am I looking at:

Q = ∫ ρ(r) dV

and if so, is it:

Q = 3rr ρ(r) 4/3πr3 dr ?

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

ahh, right, so spherical polar?

Q = π0 0 3rr ρ(r) r2sinφ dr dφ dθ ?

haruspex
Homework Helper
Gold Member
ahh, right, so spherical polar?

Q = π0 0 3rr ρ(r) r2sinφ dr dφ dθ ?
No need for that. Just answer my simple calculus question in post #4.

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 3aa 4a-r dr

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

Again, thank you!

haruspex
Homework Helper
Gold Member
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?

you mean 4πr2?

haruspex
Homework Helper
Gold Member
you mean 4πr2?
Yes!

OK,

so its...

Q = 3aa D(4a-r)⋅4πr2 dr

which is the same as...

Q = 4πD 3aa 4ar2 -r3 dr

Is this right so far?

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

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

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
Homework Helper
Gold Member
Q=(176/3)πDa4
That's what I get.

Thank you. You have the patience of a Saint!

haruspex