Equipotential surfaces electric field problem

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SUMMARY

The discussion centers on calculating the electric field magnitude and direction from equipotential surfaces. The key equation used is E = V/d, where V is the potential difference and d is the distance between equipotential lines. The participant consistently arrives at an incorrect electric field value of 2.24 x 10^2 V/m and angles of 26.6 or 63.2 degrees. The correct approach involves recognizing that the electric field is the negative gradient of the potential, which relates to the slope of the potential vs. position graph.

PREREQUISITES
  • Understanding of electric fields and equipotential surfaces
  • Familiarity with the equation E = V/d
  • Basic trigonometry for angle calculations
  • Knowledge of calculus for derivatives in electric field calculations
NEXT STEPS
  • Review the concept of electric field lines and their relationship to equipotential surfaces
  • Study the application of calculus in determining electric fields from potential functions
  • Learn how to accurately measure distances between equipotential lines
  • Explore the implications of uniform vs. non-uniform electric fields in calculations
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone seeking to understand the relationship between electric fields and equipotential surfaces.

wildredhead
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Homework Statement


A given system has the equipotential surfaces shown in the figure
What is the magnitude of the electric field?
What is the direction of the electric field? (degrees from + x axis
What is the shortest distance one can move to undergo a change in potential of 5.00

Homework Equations


E= V/d



The Attempt at a Solution


I seem to be getting the same answer and its wrong I used the pythagoreon therom to find distances of each and divided the volts by the respecitive distances. I do not know what I am doing wrong I keep getting 2.24*10^2 V/m. Moreover, when I do the angle part I just do some trig and use that. Either way I try to do it and get 26.6 or 63.2. What am I doing wrong? Is the first part just the slope
 
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Where is the figure?
 
I can't see the figure yet either, but unless you're dealing with a uniform electric field then you can't use the equation you listed. As you said, electric field is the slope of a potential vs. position graph. If you're using calculus, it's the derivative with respect to position.
 
I think I uploaded it now. Sorry!
It is an attachment
 

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Slope of the equipotential lines is tan(theta) = 2/4 = 1.2.
Find theta.
Distance between the equipotential lines d is x*sin(theta). Here x = 4 cm.
 
SO part A is the slope = -1/2? But when I do tan^-1 (2/4) I still get 26.6. What am I doing wrong?
 
What is the relationship between the directions of the electric field and the equipotential lines?
 

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