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

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Homework Help Overview

The discussion revolves around calculating the surface charge density based on a velocity over time graph. Participants are exploring the relationship between acceleration, charge, and the relevant physical constants in the context of electromagnetism.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to derive the surface charge density using various values from the graph and relevant equations. There are discussions about the correct values for constants and the implications of sign in calculations. Some participants are questioning the assumptions made regarding the constants used, particularly the permittivity of free space and the charge of an electron.

Discussion Status

There is ongoing exploration of the calculations with multiple attempts being made to arrive at the correct surface charge density. Some participants have provided guidance on the use of constants and the need for significant figures, while others are still seeking clarification on specific values and their appropriateness in the context of the problem.

Contextual Notes

Participants are noting discrepancies in the values used for calculations and the results obtained, indicating potential misunderstandings or misapplications of the relevant physics concepts. The discussion reflects a collaborative effort to clarify these points without reaching a definitive conclusion.

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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Jrlinton said:
(-2E5m/s)
The question states 2.9E5m/s

Jrlinton said:
1.6022E-19
What is th sign of sn electron's charge?
 
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?
 
TSny said:
Is the green number correct?
I have a bad habit of assuming posters know what constants to use and what their values are.
 
should this be the permittivity of free space?
 
Jrlinton said:
should this be the permittivity of free space?
I believe so.
 
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.
 
  • #10
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
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.
 
  • #12
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.
 

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