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Total charge on an insulating sphere

  • Thread starter ehilge
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  • #1
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Homework Statement



A sphere of radius R=0.0850 m is made up of insulating material and has a spherically symmetric charge distribution. The radial component of the electric field inside the sphere is given by:

Er= (-5.20E4) r3, for r ≤ R,

where Er is in N/C when r is in meters and r is the distance from the center of the sphere.

What is the total charge Q on the entire sphere? (Hint: Gauss's Law.)

Homework Equations


Gauss's law:
[tex]\oint E\bullet dA [/tex] =Qencl / epsilon0



The Attempt at a Solution



I tried to set up a triple integral in spherical coords integrating -5.2E4 [tex]\rho[/tex]3[tex]\rho[/tex] sin[tex]\phi[/tex] d[tex]\rho[/tex] d[tex]\phi[/tex]d[tex]\theta[/tex]

(btw, thats supposed to be rho cubed times rho)

from 0<=[tex]\rho[/tex]<=.085, 0<=[tex]\theta[/tex]<=2pi, and 0<=[tex]\phi[/tex]<=pi

When I calculate this and multiply by epsilon, I get: -2.56E-12
which is a power of 10 off from the given answer which is: -2.56E-11

For some reason though, I feel like this is a coincidence and I'm missing something more than a power of ten somewhere. So, does my work seem to be right? If so where am I missing a power of 10 at? And also, I'm thinking there is an easier way to do this problem other than integrating so any insight into that would be great!
Thanks!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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I think it's interesting that you chose to use a triple integral. Here's what I would do:

To get the q(inside) all you have to do is integrate E * 4 pi r^2 * epsilon naught * dr. Remember to get the area in terms of dr now and not dA. Try it and see what you get!

**Edit** I've been trying to solve it now with your numbers, and am not getting the value that you listed above (-2.56E-11). At one point I get (-2.56 *10^-17), but that isn't close. Are all the numbers you provided correct?
 
Last edited:
  • #3
Doc Al
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To get the q(inside) all you have to do is integrate E * 4 pi r^2 * epsilon naught * dr. Remember to get the area in terms of dr now and not dA.
The flux is just E*Area = E(at r = R)*4 pi R^2.
 
  • #4
163
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The flux is just E*Area = E(at r = R)*4 pi R^2.
That did it and is definitely easier than the way that I was trying to compute the answer, thanks!

And ltjlogic, your idea was what I had actually tried first. I'm not entirely sure why it doesn't work but I'm assuming it has something to with the fact the its measuring the charge on a line from the center of the sphere to r. Phi and theta are needed to complete the entire sphere. Not entirely sure on this reasoning though. Thanks for your suggestion.
 

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