The maximum magnitude of the electric field of a sphere

[sliperyfrog]

Homework Statement

A non-conducting sphere of radius a carries a non-uniform charge density. The electrostatic field inside the sphere is a distance b from is center and is given by E = (b/a)⁴E₀ (E₀ being the maximum magnitude of the field.)

a. Find an expression for E₀ in terms of the total charge Q₀ and the radius of the sphere.

b. Determine the charge density of the sphere as a function of radius.

Homework Equations

ε₀ ∫ E ⋅ dA = ∫ ρdV

dQ = ρdV[/B]

The Attempt at a Solution

[/B]
So for a I did ε₀ ∫ E ⋅ dA = ε₀E4πa² since
E = (b/a)⁴E₀
ε₀E4πa² = ε₀(b/a)⁴E₀)4πa² = 4πE₀ε₀(b⁴/a²)

For the ∫ ρdV side of the problem my professor did it a problem similar in his slides for b > a of uniform charge density the ∫ ρdV = Q₀ so i just assume it is the same for a non-uniform charge density. Getting me
4πE₀ε₀(b⁴/a²) = Q₀
thus E₀ = (a²Q₀)/(4πb⁴)

For part b I did dQ₀ = ρdV = ρ4πa²da so
Q₀ = ∫₀^b ρ4πa²da = (4/3)πρb^3 so
ρ = (3Q₀)/(4πb³)

 

[haruspex]

i just assume it is the same for a non-uniform charge density

You do not need to assume this. The field you are given is spherically symmetric, so you know that if you decompose it as nested thin spherical shells each shell is uniformly charged. What do you know about the field outside such a shell? You can go straight from that to the answer.

For part b I did dQ0 = ρdV = ρ4πa2da

I don't follow your thinking here. Q₀ is a constant, so dQ₀ means nothing (or zero).

 

[TJGilb]

You seem to have some misconceptions, so let's just start at the beginning. Take a look at your equation for E and at what radius it will be at its maximum of E₀. You should notice that when a=b, you'll have E=E₀. What does this tell you about the charge enclosed by your Gaussian surface when you have maximum E?

 

[haruspex]

Q0 = ∫0b ρ4πa2da

It would make more sense to write that as $Q(b)=\int_0^b\rho\pi r^2.dr$, but

= (4/3)πρb^3

No, to get that ρ would have to be constant.
Instead, can you find two things that should equate to $\frac{dQ}{db}$?

 

[sliperyfrog]

I don't follow your thinking here. Q0 is a constant, so dQ0 means nothing (or zero).

That is typo I meant that when you integrate dQ is comes out to the total charge of Q which is Q₀

No, to get that ρ would have to be constant.
Instead, can you find two things that should equate to dQdb\frac{dQ}{db}?

I am confused on why i would need to find dQ/db or what dQ/db is supposed to be.

You seem to have some misconceptions, so let's just start at the beginning. Take a look at your equation for E and at what radius it will be at its maximum of E0. You should notice that when a=b, you'll have E=E0. What does this tell you about the charge enclosed by your Gaussian surface when you have maximum E?

That when the sphere is at the maximum of E₀ the electrostatic sphere is a distance a from the center of the sphere and the charge would be at its maximum.

 

[TJGilb]

Right, the charge is at its maximum. So while using Gauss' Law, the q_enc is simply Q₀, and the radius of your Gaussian surface is the radius of the sphere. That gives you everything you need to put E₀ in terms of Q₀ and a, no integration needed.

 

[TJGilb]

Relooking at what you wrote (it seems clearer now for some reason), you basically have it, you just need to plug in a for b. Then plug what you get for E₀ into E, plug that back into Gauss' Law, and you have the result of the charge density integral. So if you have the result of the integral (which is with respect to b) you can find the charge density by taking the derivative of that result with respect to b.

 

[haruspex]

what dQ/db is supposed to be.

I wrote Q=Q(b), meaning the total charge inside radius b. See my first line in post #4.

 

[sliperyfrog]

Relooking at what you wrote (it seems clearer now for some reason), you basically have it, you just need to plug in a for b. Then plug what you get for E0 into E, plug that back into Gauss' Law, and you have the result of the charge density integral. So if you have the result of the integral (which is with respect to b) you can find the charge density by taking the derivative of that result with respect to b.

So what you are sayings is I take my answer for a E₀ = (a²Q₀)/4πb⁴ε₀ since b=a when E₀ = E I replace b for a so E₀ = Q₀/4πa²ε₀ Then i take that answer for a and plug it into E so I get E = (Q₀b⁴)/4πa⁶ε₀ Than i plug that into guass's law so ε₀∫E⋅dA = (ε₀4πa²Q₀b⁴)/4πa⁶ε₀ = (Q₀b⁴)/a⁴ so then (Q₀b⁴)/a⁴ = ∫ρdV I then take the derivative of both sides and i get 3Q₀b³/a⁴ = (ρ4πb³)/3 so ρ = (9Q₀)/4πa⁴

 

[TJGilb]

Almost. It looks like you're plugging it back into Gauss' law incorrectly. Since E has no angular variance, when you plug it in you should get $\mathbf E \int_S dA$ which, with the integral of the Gaussian surface being that of a sphere, should come out to $\frac {Q_0 b^4} {4 \pi a^6 \epsilon_0} 4 \pi b^2 = \frac 1 {\epsilon_0} \int_V \rho dV$.

 

[sliperyfrog]

Why when i take the surface integral of dA would it come out to 4πb² instead of 4πa² isn't the area supposed to be base of the radius of the sphere?

 

[TJGilb]

It's the radius of your Gaussian Surface, which in this case is any arbitrary radius equal to or less than a (since you're only interested in values of the sphere to find the charge density, and your equation for the electric field is only good for those values anyways).

 

[sliperyfrog]

So would I do (Q₀b⁶)/a⁶ = ∫ρdV then i take the derivative of both sides so (6Q₀b⁵)/a⁶ = (ρ4πb³)/3 so
ρ = (9Q₀b³)/2a⁶

 

[TJGilb]

Yes, but when you take the derivative of the integral, you're basically just canceling it out.

 

[sliperyfrog]

I just realized I keeping trying to turn ∫dV into V like i am taking the integral. So it would just be
ρ = (6Q₀b⁵)/a⁶

 

[TJGilb]

Yep, good job!

 

[sliperyfrog]

Thanks for the help!