Electric field from potential energy

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SUMMARY

The discussion focuses on calculating the electric field from potential energy in a uniform electric field scenario. The potential at points (0,0) and (.5,0) is 20V higher than at (0,.5), leading to the conclusion that the electric field has a magnitude of 40 N/C in the positive j-direction. The reasoning is based on the relationship between electric potential difference and electric field, confirming that the field points in the direction of decreasing potential. The correct answer is option b) 40 N/C j.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with the relationship between potential difference and electric field
  • Knowledge of vector components in Cartesian coordinates
  • Basic principles of electrostatics
NEXT STEPS
  • Study the concept of electric potential difference in uniform electric fields
  • Learn about the mathematical derivation of electric field from potential energy
  • Explore the implications of electric field direction in relation to potential
  • Review problems involving electric fields and potentials for exam preparation
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Students preparing for exams in physics, particularly those focusing on electrostatics and electric fields, as well as educators teaching these concepts.

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Homework Statement



Consider uniform electric field. The values of the potentials at the points(0,0) (.5,0) are equal and 20V higher than the potential at (0,.5) The magnitude and direction of the electric field are:
a)0
b)40N/c j
c)-40N/c j


Homework Equations





The Attempt at a Solution


I know the answer is b because this is a problem for my exam review and we were given the solutions, I don't know how to get it though can anyone help?
 
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I think I know why. I am thinking that since the field points in the direction of decreasing potential it must go up and since Va-Vb = E(b)-E(a) you know the potential difference is 20. so 20=E(.5) (a=0) then divide out and get 40n/c j. Also you know that the field must be perpendicular to the x-axis because as you move down the x-axis the potential remains constant right? Can anyone verify this I have an exam tomorrow and I want to make sure I have the concepts right. Thanks.
 
You're reasoning is correct.
 

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