Determining Work Based on a Charge and Particle

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Homework Help Overview

The discussion revolves around calculating the work done in bringing a charge from infinity to a specific point in an electric potential field, with a given potential of -4.5E3 V and a charge of 0.94 C.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between voltage, charge, and work, with attempts to apply the formula for calculating work based on electric potential. There are questions regarding the correctness of unit conversions and the final result.

Discussion Status

Some participants express uncertainty about the calculations and whether the results align with expected outcomes. There is acknowledgment of a potential typo in the calculations, but the overall direction of the discussion appears to validate the approach taken.

Contextual Notes

Participants mention concerns about the nature of the problem, indicating that the instructor may present trick questions, which adds to their hesitation regarding the simplicity of the solution.

rabiddogma
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Homework Statement


The potential at a point is -4.5E3 V. How much work, in J, is done to bring a 0.94 C charge from infinity to the point?

Homework Equations


1V = 1 J/C

The Attempt at a Solution


Well I know that 1 V = 1 J/C so I did -4500V*0.94C = -4230J but I wasn't sure if this was right.
 
Last edited:
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rabiddogma said:

Homework Statement


The potential at a point is -4.5E3 V. How much work, in J, is done to bring a 0.94 C charge from infinity to the point?


Homework Equations


1V = 1 J/C


The Attempt at a Solution


Well I know that 1 V = 1 J/C so I did -4500V/0.94C = -4230J but I wasn't sure if this was right.

Check how the units multiply out in your calculation. Do you get Joules?
 
gneill said:
Check how the units multiply out in your calculation. Do you get Joules?

Ah my bad, that was a typo, what I did was -4500V*0.94C. Which does come out to J but I wasn't sure if that gave the proper answer.
 
rabiddogma said:
Ah my bad, that was a typo, what I did was -4500V*0.94C. Which does come out to J but I wasn't sure if that gave the proper answer.

Alright. In that case the result looks fine :smile:
 
gneill said:
Alright. In that case the result looks fine :smile:

Alright, thanks, I just wasn't sure because my instructor has a tendency of giving us trick problems and that just seemed too easy.
 

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