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Gaussian Surface Derivations

  • Thread starter nealh149
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  • #1
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For the following Eo is epsillon zero



Homework Statement



41. A solid nonconducting spere of radius R has a uniform charge distribution of volume charge density p = kr/R where k is constant and r is the distance from the center. Show the (a) the total charge on the sphere is Q = pikR^3. (I did this, it's fine) and (b) that

E = (1/(4piEo)(Q0R^4)(r^2) gives the magnitude of the electric field inside the sphere.

The Attempt at a Solution




41. For 41 (b) I tried to set up an integral. I new that Q = pV

Thus: Q = (kr/R)(4/3)(pi)(r^2) = [4kr^3(pi)}/[3R]

Because the gaussian surface is spherically symmetric

E = Q/(Eo*A) E = {[4kr^3(pi)}/[3R]} / (Eo*(4(pi)r^2)

This reduces to (kr)/(3EoR)

My proof seems to work alright, but it's not what the book asks you to find.
 

Answers and Replies

  • #2
Doc Al
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41. For 41 (b) I tried to set up an integral. I new that Q = pV

Thus: Q = (kr/R)(4/3)(pi)(r^2) = [4kr^3(pi)}/[3R]
To find the charge, you need to integrate p dV. What you did was treat p as a constant--it's not constant, it's a function of r.
 
  • #3
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I tried to do it a totally different way, without any integration.

We know because density is a funciton of r we can set up a ratio.

Total Charge / Total Volume = Enclosed Charge / Enclosed Volume

Total Charge * Enclosed Volume / Total Volume = Enclosed Charge

So (pikR^3)(4pir^2) / (4piR^2) Q enclosed = [pi][k][r^3]

So put this into the Electric Field Formula

EA = Q/Eo So E = Q/AEo So E = ([pi][k][r^3])/(4pir^2)/Eo

So E = kr/4Eo

This seems logically correct, but not what there looking for. Can somebody give me some hints on how to do this correctly?
 
  • #4
Doc Al
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I tried to do it a totally different way, without any integration.
You must integrate!

We know because density is a funciton of r we can set up a ratio.

Total Charge / Total Volume = Enclosed Charge / Enclosed Volume
This would be true if the charge density were constant, but it's not.

Instead, integrate:

[tex]Q = \int \rho dV = \int \frac{kr}{R} 4 \pi r^2 dr[/tex]
 
  • #5
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[tex]Q = \int \rho dV = \int \frac{kr}{R} 4 \pi r^2 dr = \frac{kr}\int{R}4 \pi r^2[/tex]

Thus Q = (4kpiR^3)/3 plugged into the gauss formula from before you get kR/3Eo

Still not working. I'm really having trouble
 
  • #6
Doc Al
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Redo that integral to get Q. I wrote it so you can see where each term originates--to do the actual integration, combine the r factors.
 
  • #7
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Ok I got (kr^2)/(4REo) Still Different.
 
  • #8
Doc Al
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To put the answer in the form they want, eliminate k. Use the formula you derived for total Q (in your first post).
 

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