Equipotential Surfaces and Electric Field

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SUMMARY

The discussion focuses on calculating the electric field magnitude and direction from given equipotential surfaces. The electric field magnitude was determined to be 559 V/m using the equation E = -V/R, where R is calculated as 4sin(26.565). The user initially struggled with determining the correct angle for the electric field direction but ultimately resolved the issue. The change in potential for a distance of 1.50 V was also addressed, indicating a comprehensive understanding of the topic.

PREREQUISITES
  • Understanding of electric fields and equipotential surfaces
  • Familiarity with the equation E = -V/R
  • Knowledge of trigonometric functions, specifically sine
  • Ability to interpret graphical representations of electric fields
NEXT STEPS
  • Explore the relationship between electric field strength and potential difference
  • Learn about the concept of equipotential lines in electrostatics
  • Study the implications of electric field direction in circuit analysis
  • Investigate the applications of electric fields in real-world scenarios
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric fields and equipotential surfaces.

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Homework Statement



A given system has the equipotential surfaces shown in the figure .
http://session.masteringphysics.com/problemAsset/1122530/1/Walker.20.39.jpg

A)What is the magnitude of the electric field?
B)What is the direction of the electric field? (in degrees from the +x axis)
C)What is the shortest distance one can move to undergo a change in potential of 1.50 V?


Homework Equations


E=-V/R


The Attempt at a Solution


I found the answer to the Part A:
E= -V/R, R=4sin(26.565)
E=559 V/M

I cannot wrap my head around the answer to Part B, though. I thought it would simply be 180-25.65, but that's wrong. Any help?

I also know how to do Part C, so just Part B would be great!
 
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Answers that are also not correct are:
153
63.4
117.
 
Nevermind, I was doing the angle perpendicular in the wrong direction. I got the answer.
 

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