How Is Electric Potential Difference Calculated?

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Homework Help Overview

The discussion revolves around calculating electric potential differences in the context of electric fields, specifically focusing on the work done on an electron as it moves between points A, B, and C. Participants are exploring the relevant formulas and concepts related to electric potential.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate formulas for calculating electric potential differences, questioning the relevance of different equations provided. There is also an exploration of the concept of equipotential lines and the implications for potential energy.

Discussion Status

Some participants have confirmed initial thoughts regarding the calculations and the relationship between work and potential difference. There is an acknowledgment of the negative sign in the equations, and one participant has reported a calculated value for the potential difference, although the discussion remains open-ended without a definitive conclusion.

Contextual Notes

Participants are navigating the definitions and implications of electric potential, including the concept of path independence in electric fields. There is a mention of the charge of the electron as a critical factor in the calculations.

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[Solved] Electric Potential Differences

When an electron moves from A to B along an electric field line in Fig. 25-26, the electric field does 2.44*10^-19 J of work on it.

http://tinyurl.com/2s34zk

Find the electric potential differences from VB->VA and VC->VA.

I think the potential difference is the same for each but I'm not positive. I'm not sure what formula to use either...V2-V1= -Welec / q or V2-V1= - integral E*ds ? If someone could get me started in the right direction I think I can figure out the rest. Thanks for your help!
 
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Your guess is correct for the first part. The reason why(which is important to understand!)is in the names. If you move in the direction of an equipotential line, you're not changing your potential energy, and hence there's no work being done, in going from one equipotential to another, you change your potential energy by that much, no matter the path(ever hear that electric fields are path indepedent?)

As for the second part, I think both equations are true, but only one is of any actual use to you. Which is it, considering what information was given?
 
I think that the only thing you need to do is divide the work by the charge of electron, since potential is defined as work done on one coulomb charge. the potentials between AB and AC are the same.
 
Thanks for confirming my suspicions. The negative sign was throwing me off. Found the electrical potential difference for both parts to be 1.525V. Thanks again!
 

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