Surface Charge Density

  • Thread starter Jrlinton
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  • #1
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Homework Statement


upload_2017-2-7_10-37-7.png


Homework Equations


E=σ/(2Eo)
σ=2Eoma/e
a=Δv/Δt

The Attempt at a Solution


So doing my best to read the velocity over time graph I came up with
Δv/Δt=(-2E5m/s)/(10E-12s)=-2E17 m/s/s

σ=2Eoma/e
=2(8.99E-9C)(9.109E-31kg)(-2E17m/s/s)/(1.6022E-19)
=-0.020444 C/m^2
but by the first illustration you can see that the sheet has a positive charge so the charge density should be:
0.020444 C/m^2 <----this final answer was incorrect
 

Answers and Replies

  • #3
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Okay let's try this again.
A=(2*-2.9E5m/s)/(2E-11s)
=-2.9E16
So using that correct value along with the appropriate value for the charge of an electron:
2(8.99E9C)(9.109E-31kg)(-2.9E16)/(-1.6022E-19)
2.964E15 C/m^2
 
  • #4
haruspex
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Okay let's try this again.
A=(2*-2.9E5m/s)/(2E-11s)
=-2.9E16
So using that correct value along with the appropriate value for the charge of an electron:
2(8.99E9C)(9.109E-31kg)(-2.9E16)/(-1.6022E-19)
2.964E15 C/m^2
Looks ok to me now. Still marked as wrong?
 
  • #5
TSny
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So using that correct value along with the appropriate value for the charge of an electron:
2(8.99E9C)(9.109E-31kg)(-2.9E16)/(-1.6022E-19)
2.964E15 C/m^2
Is the green number correct?
 
  • #6
haruspex
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Is the green number correct?
I have a bad habit of assuming posters know what constants to use and what their values are.
 
  • #7
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should this be the permittivity of free space?
 
  • #9
TSny
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I have a bad habit of assuming posters know what constants to use and what their values are.
I would have overlooked it also, but his/her result for σ was humongous.
 
  • #10
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Let's try this again
The acceleration according to the graph is 2*(-2.9E5m/s)/2E-11= -2.9E16 m/s/s

σ=2Eoma/e
=2*(8.855E-12)(9.109E-31)(-2.9E16)/(-1.6022E-19)
=3E-6 C/m^2
Still not acceptable
 
  • #11
haruspex
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Let's try this again
The acceleration according to the graph is 2*(-2.9E5m/s)/2E-11= -2.9E16 m/s/s

σ=2Eoma/e
=2*(8.855E-12)(9.109E-31)(-2.9E16)/(-1.6022E-19)
=3E-6 C/m^2
Still not acceptable
You can either use Coulomb's constant, but as a divisor here, or the permittivity as a multiplier. But in the second you need a 4π too.
Edit: no that's wrong... Let me check.

Edit2: I should have written, you can either divide by Coulomb's constant and by 4π, or multiply by the permittivity, as you did. Anyway, TSny seems to have identified the remaining issue.
 
  • #12
TSny
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Let's try this again
The acceleration according to the graph is 2*(-2.9E5m/s)/2E-11= -2.9E16 m/s/s

σ=2Eoma/e
=2*(8.855E-12)(9.109E-31)(-2.9E16)/(-1.6022E-19)
=3E-6 C/m^2
Still not acceptable
Looks good. But the data was given to 3 significant figures.
 

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