Calculating work done by an electric field

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Homework Help Overview

The original poster is attempting to calculate the work done by an electric field in moving a proton between two points with different electric potentials, specifically from +155V to -75V. The problem involves understanding the relationship between charge, potential difference, and work done.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equation for work done by an electric field and explore how to incorporate the change in potential into their calculations. There is a focus on clarifying the correct expression for potential difference and its implications for the work calculation.

Discussion Status

Some participants have provided guidance on the equation to use and have engaged in verifying calculations. There is ongoing exploration of the correct change in potential, with some participants questioning earlier calculations and suggesting corrections.

Contextual Notes

Participants are discussing the implications of potential differences and ensuring that the correct values are used in calculations. There is a focus on understanding the signs and values associated with electric potential.

crh
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Homework Statement



How much work does the electric field do in moving a proton from a point with a potential of +155V to a point where it is -75V. Express your answer in both joules and electron volts.

Homework Equations



W = -qV

The Attempt at a Solution



I know that q=(1.6E-19) but I am not for sure how to go about incorporating my two potentials. Can someone give me some help? I thank you in advance!
 
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crh said:

Homework Equations



W = -qV
Perhaps it would help if the equation was written thus:

[tex]W = -q\Delta V[/tex]

In other words, the work done by the electric field on the charged particle is the negative product of the charge and the change in potential.
 
Ok I think I figured it out. Tell me if I am wrong.

W = -qV, but V is potential E (PE), so therefore V = (PE2-PE1)

so...

W = -(1.60E-19C)(-80V) = 1.28E-17 J
and
= -(1e)(-80V) = 80eV

did I go about this right?
 
crh said:
Ok I think I figured it out. Tell me if I am wrong.

W = -qV, but V is potential E (PE), so therefore V = (PE2-PE1)

so...

W = -(1.60E-19C)(-80V) = 1.28E-17 J
and
= -(1e)(-80V) = 80eV

did I go about this right?
You're on the right lines but be careful when calculating the change in potential.
 
Are you meaning scientific notation?
 
crh said:
Are you meaning scientific notation?
Nope, something somewhat simpler than that. What is the difference between -75 and +155?
 
oh it needs to be -230V.
 
crh said:
oh it needs to be -230V.
Much better :approve:
 

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