What is the Surface Charge Density Based on Velocity Over Time Graph?

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Jrlinton
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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
 
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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
 
Jrlinton said:
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?
 
Jrlinton said:
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?
 
should this be the permittivity of free space?
 
haruspex said:
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.
 
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
 
Jrlinton said:
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.
 
Jrlinton said:
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.