The Electric Potential Equation

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SUMMARY

The discussion centers on calculating the magnitude of the electric field given a charge of q = −4.00×10−9 C that moves 0.500 m in a uniform electric field, resulting in a kinetic energy of 5.00×10−7 J. The relevant equations include U = kQ/r², F = Eq, W = Fd, and U = Eqd. The user correctly identifies that the voltage difference can be expressed as the product of electric field, charge, and distance, leading to the conclusion that the electric potential at point B is 155 V, while the potential at point A is +30.0 V.

PREREQUISITES
  • Understanding of electric potential and electric fields
  • Familiarity with the concepts of kinetic energy and work
  • Knowledge of the equations U = kQ/r² and W = Fd
  • Basic principles of charge movement in electric fields
NEXT STEPS
  • Study the relationship between electric potential and electric field strength
  • Learn how to apply the work-energy theorem in electric fields
  • Explore the derivation of the electric potential equation from fundamental principles
  • Investigate the implications of charge movement in varying electric fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding electric potential and field interactions.

meaghan
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Homework Statement



An object with charge q = −4.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. After the charge has moved to point B, 0.500 m to the right, it has kinetic energy 5.00×10−7 J .What is the magnitude of the electric field?

Homework Equations


U = kQ/r^2
F = Eq
W = Fd
U = Eqd

The Attempt at a Solution


I was thinking that I could maybe set the voltage difference equal to the electric field * charge * distance since the part before asked that if the electric potential at point A is +30.0 V, what is the electric potential at point B? Which the answer was 155 V.
 
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meaghan said:
I was thinking that I could maybe set the voltage difference equal to the electric field * charge * distance since the part before asked that if the electric potential at point A is +30.0 V, what is the electric potential at point B? Which the answer was 155 V.
By what logic, theory, or standard equation is a potential equal to field x charge x distance?
 

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