Gauss's Law (i think)

  • Thread starter physizl
  • Start date
  • #1
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Homework Statement



Determine the magnitude of the electric field at the surface of a lead-199 nucleus, which contains 82 protons and 117 neutrons. Assume the lead nucleus has a volume 199 times that of one proton and consider a proton to be a sphere of radius 1.20 10-15 m.


Homework Equations



E = kQ/r^2
Q = pV
p = Ne = (#protons)(1.6x10^-19)
V = Volume Nucleus = (#lead)(4/3)(pi)(r^2)

The Attempt at a Solution



tried plugging all the numbers in and came out with a wrong answer..

is my method correct?
is there an easier way such as enclosing the atom in a Gaussian sphere?
..making E = sigma/epsilon_naught
what would sigma be in this case?
 

Answers and Replies

  • #2
318
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Welcome to the forum.

Why do you take Q= pV ?
What did you take for r? In the Electric field?
 
  • #3
9
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Q = enclosed charge?
p = charge density & V = volume, so the units are just C when you multiply pV

for r, i set whatever the volume is and i set it to equal (4/3)pi(R^3) and solved for R
 
  • #4
9
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nevermind i figured how to do it!

i used..
E = kQ/(r^2)
Q = (# protons)(elementary charge)
r = [(# lead)^(1/3)](radius proton)

plugged Q and r back in E and came out with the correct answer
 

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