Find the electric field at an arbitrary point

  • #1
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Homework Statement



A distribution of charge with spherical symmetry has volumetric density given by: $$ \rho(r) = \rho_0 e^{ \frac {-r} {a} }, \left( 0 \leq r < \infty \right); $$
where ##\rho_0## and ##a## is constant.

a) Find the total charge
b) Find ##\vec E## in an arbitrary point

Homework Equations



I've already found the answer of a): ## Q_t = 8\pi a^3 \rho_0##

The Attempt at a Solution



To solve b), I've used the Gauss law (spherical symmetry) and that's what I've found $$ \vec E = \frac {4\pi a^3 \rho_0} {r^2} \vec r .$$
This answer seems to me very acceptable, but I've looked at the solutionary and there's another result pretty much complicated. What do you think?
 

Answers and Replies

  • #2
vela
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That's wrong.
 
  • #3
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To find a total charge you need to integrate the density ##\rho## over whole space (spherical coordinates are best here). Luckily ##\rho## gets small pretty quickly so integration gives finite result. a) is exactly like finding a mass of an infinite object with density ##\rho##.
 
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  • #4
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That's wrong.
Which one is wrong?

To find a total charge you need to integrate the density ##\rho## over whole space (spherical coordinates are best here). Luckily ##\rho## gets small pretty quickly so integration gives finite result. a) is exactly like finding a mass of an infinite object with density ##\rho##.
Yes, that's what I did at the answer of a)... Is that answer wrong? I'm not sure if b) answer is right.
 
  • #5
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Yes, that's what I did at the answer of a)... Is that answer wrong?
No, sorry I didn't see that in "The attempt at a solution". a) is correct.
 
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  • #6
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That's wrong.
Would you please help me to get the right answer for ##\vec E##?
 
  • #7
haruspex
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Would you please help me to get the right answer for ##\vec E##?
Please show your working for part b. Without that it is hard to say where you are going wrong.
It does seem clear to me that there should be an exponential in the answer.
 
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  • #8
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Please show your working for part b. Without that it is hard to say where you are going wrong.
It does seem clear to me that there should be an exponential in the answer.
First I've used the Gauss law, with the information I got from a): $$ E r^2 4 \pi = \frac {8 \pi a^3 \rho_0} {\varepsilon_0 r^2} \\ \vec E = \frac {4 \pi a^3 \rho_0} {\varepsilon_0 r^2} \vec r .$$
The integral of the left side I did under spherical cordinates and the right side I've used the answer of a).
 
  • #9
haruspex
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First I've used the Gauss law, with the information I got from a): $$ E r^2 4 \pi = \frac {8 \pi a^3 \rho_0} {\varepsilon_0 r^2}$$
The information you show from (a) was the total charge across an infinite region. r therefore does not appear in that expression. I do not see how you can get the above by applying Gauss' law to that.
What do you get for the charge inside radius r?
 
  • #10
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The information you show from (a) was the total charge across an infinite region. r therefore does not appear in that expression. I do not see how you can get the above by applying Gauss' law to that.
What do you get for the charge inside radius r?
Oh thank you, I was doing my calculation wrong. So I've tried to do the integrals over r and I got a different answer this time (it doesn't match with the solutionary): $$ E \int_0^r \, da = \frac {4 \pi \rho_0} {\varepsilon_0} \int_0^r e^{\frac {-r} {a}} \, dr \\ 4 \pi r^2 E = \frac {4 \pi \rho_0} {\varepsilon_0}\left. \left(-a e^{\frac {-r} {a}} \right)\right|_0^r \\ \vec E = \frac { \rho_0} {\varepsilon_0 r^2} \left( a - e^{\frac {-r} {a}} \right) \vec r $$
 
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  • #11
haruspex
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Oh thank you, I was doing my calculation wrong. So I've tried to do the integrals over r and I got a different answer this time (it doesn't match with the solutionary): $$ E \int_0^r \, da = \frac {4 \pi \rho_0} {\varepsilon_0} \int_0^r e^{\frac {-r} {a}} \, dx \\ 4 \pi r^2 E = \frac {4 \pi \rho_0} {\varepsilon_0}\left. \left(-a e^{\frac {-r} {a}} \right)\right|_0^r \\ \vec E = \frac { \rho_0} {\varepsilon_0 r^2} \left( a - e^{\frac {-r} {a}} \right) \vec r $$
You are still going wrong right at the start.
The charge density is ##\rho(r)=\rho_0e^{-\frac ra}##.
Consider a spherical shell, radius r, thickness dr. What is the charge on the shell?
 
  • #12
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I've took out ##\rho_0## of the integral because it's a constant...
 
  • #13
haruspex
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I've took out ##\rho_0## of the integral because it's a constant...
That does not explain it.
I ask again, what is the charge on the shell? It is not ρ - that is the charge density on the shell.
 

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