Electric Field of a Non-Uniformly Charged Sphere | Homework Help

[Dgeld]

Homework Statement

A solid isolated sphere with radius R has a non uniform charge which is given by ρ= Ar²,
with A a constant and r<R measured from the centre of the sphere

Homework Equations

(a) Show that the electric field outside the sphere is equal to E = (AR⁵)/(5ε₀r²)
(b) Show that the electric field inside the sphere is equal to E = (AR³)/5ε₀)

The Attempt at a Solution

No attempt, I am completely clueless about what to do here.
Pardon my english, not a native speaker.

 

[BvU]

Hello Dg,

Your english is just fine and certainly good enough to read the guidelines . There you will find that we can't help unless you make an effort.

As a welcoming gesture: did you notice the charge density is spherically symmetrical ? And does that remind you of a useful theorem named after a long-dead german genius ?

 

[Dgeld]

Hello Dg,

Your english is just fine and certainly good enough to read the guidelines . There you will find that we can't help unless you make an effort.

As a welcoming gesture: did you notice the charge density is spherically symmetrical ? And does that remind you of a useful theorem named after a long-dead german genius ?

Yes i know it's somthing to do with Gausses law, Φ = ∫ EdA which i can then change into Q = ρ E^DV
I can fill in ρ as the Q/ Volume of sphere and dV for a surface area. But than i am stuck with an integral i do not know how to solve.

 

[BvU]

Write down the complete expression for that integral (counts as attempt at solution !)

 

[Dgeld]

Well the full expression then would be Q = ∫^r₀ (q/πr4)r(4πr²dr)

My apologies, i do not know how to put a combination of sub and super script on an integral, the 0 is supposed to be at the bottom.

 

[BvU]

yeah, the ordinary fonts aren't very good at that. To do it right you need $LaTeX$
I interpret your formula as $$ Q = \int_0^R {q\over 4\pi r} r \; 4\pi r^2 \,dr $$ correct me if I am wrong (i.e. for example: upper limit R not r)

Can you explain what your Q stands for and how you set it up to look like this ? What is lower case q ?

And we are talking about part (a) with $r > R$, right ?

With Gauss' law in this spherically symmetric case, you can write $E = \displaystyle {Q\over 4\pi\varepsilon_0 \; r^2}\$ with $Q$ the total charge; agree ?Also: re exercise shrewdness : if you see a 5th power of R divided by 5 in the answer desired, doesn't that ring a bell about a possible integrand ?
$\mathstrut$

 

[Dgeld]

Yes about part (a).

the lower case q should be uppercase, my mistake. The Q stands for the charge of the whole sphere.

And yes i agree with that part, But i don't think i can fill in Q by what i have managed to get to find E or can i?

Also, i suppose it doesn't ring a bell for me as i have no idea how that could lead me to a possible integrand.

Oh and btw, thank you very much for helping me!

 

[BvU]

You are welcome.

If your q is actually Q then you have a circular reference in excel language : the variable is both on the left and on the right side of the equation

The expression to get Q when given $\ \rho = Ar^2\$ is what we are still struggling with, right ?

If, in a small volume $dV\,$ the charge density is $\rho, \$ then the charge in that small volume is $\rho dV$.

For a larger volume $V$ with a charge density $\rho(\vec r)$ (depending on position), the total charge is then $Q=\int_V \rho(\vec r) \, dV$.

Can you now write the correct version of your integral in #5 ? You have all the ingredients!