Electric field question

  • #1
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Homework Statement


A solid isolated sphere with radius R has a non uniform charge wich 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, im completely clueless about what to do here.
Pardon my english, not a native speaker.
 

Answers and Replies

  • #2
BvU
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Hello Dg, :welcome:

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 :rolleyes: ? And does that remind you of a useful theorem named after a long-dead german genius :wink: ?
 
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  • #3
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Hello Dg, :welcome:

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 :rolleyes: ? And does that remind you of a useful theorem named after a long-dead german genius :wink: ?
Yes i know it's somthing to do with Gausses law, Φ = ∫ EdA wich i can then change into Q = ρ EDV
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.
 
  • #4
BvU
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Write down the complete expression for that integral (counts as attempt at solution !)
 
  • #5
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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.
 
  • #6
BvU
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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 excercise shrewdness :wink: : 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##
 
  • #7
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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 dont think i can fill in Q by what i have managed to get to find E or can i?

Also, i suppose it doesnt 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!
 
  • #8
BvU
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You are welcome.

If your q is actually Q then you have a circular reference in excel language :biggrin: : 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!
 

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