Find the change in electric potential

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SUMMARY

The discussion centers on calculating the change in electric potential in a uniform electric field of 6.0 x 10^5 N/C directed along the positive x-axis. The user incorrectly applies the formula ∆V = -E*∆s, assuming a change in potential due to movement in the y-direction. The correct conclusion is that the change in electric potential between the origin and the point (0, 6.0m) is zero, as movement in the y-direction does not affect potential in a uniform electric field aligned with the x-axis.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the formula ∆V = -E*∆s
  • Knowledge of equipotential surfaces
  • Basic principles of electromagnetism
NEXT STEPS
  • Study the concept of equipotential surfaces in electric fields
  • Learn about uniform electric fields and their properties
  • Explore the implications of electric potential in different coordinate directions
  • Review examples of potential difference calculations in various electric field configurations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to deepen their understanding of electric fields and potential differences.

Brit412
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Homework Statement



I'm having trouble understand why I'm getting the following question wrong: a uniform electric field of magnitude of 6.0*10^5 N/C points in the positive x direction. Part A) Find the change in electric potential between the origin and the point (0, 6.0m)

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Homework Equations



Wouldn't it just be ∆V = -E*∆s = (6.0*10^5) (6.0m)?.

The Attempt at a Solution

 
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If the field is uniform in x direction and you want the voltage change from 0,0 to 0,6 it's 0. anything in the Y direction will be equipotential.
 

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