Electric Field & Point Charge Potential Energy

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SUMMARY

The discussion focuses on calculating the change in electric potential energy (\(\Delta U\)) for an 8.0 µC charge moving through a uniform electric field of 8.5×105 N/C. The initial incorrect approach used the formula \(\Delta U = q\Delta V\), leading to a miscalculation of 1.13 J. The correct formula, \(\Delta U = q_0 E d\), was applied to yield the accurate result of -40.8 J, confirming that the charge moves against the electric field.

PREREQUISITES
  • Understanding of electric fields and potential energy
  • Familiarity with the formula \(\Delta U = q\Delta V\)
  • Knowledge of the relationship between electric field strength and voltage (\(E = -\Delta V/\Delta s\))
  • Basic principles of charge movement in electric fields
NEXT STEPS
  • Study the derivation and applications of \(\Delta U = q_0 E d\)
  • Learn about electric field lines and their impact on charge movement
  • Explore the concept of electric potential difference and its calculations
  • Investigate the implications of electric potential energy in circuit design
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone involved in electrical engineering or related fields seeking to deepen their understanding of electric potential energy and fields.

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Homework Statement


A uniform electric field of magnitude 8.5×105 N/C points in the positive x direction.

Find the change in electric potential energy of a 8.0 \muC
charge as it moves from the origin to the point (6.0 m , 0).

Homework Equations



1 N/C = 1 V/m
\DeltaU = q\DeltaV
E = -\DeltaV/\Deltas

The Attempt at a Solution



8.5x105 N/C = 8.5x105 V/m

(8.5x105 V/m) / (6m) = 1.41x105V <----> E = -\DeltaV/\Deltas

1.41x105 V * 8.0x10-6C = 1.13J <-----> \DeltaU = q\DeltaV

(the correct answer is -41J; I'm trying to figure out how to get there)
 
Last edited:
Physics news on Phys.org
Wrong equation used:

\DeltaU=q0Ed

\DeltaU=8x10-6C* -8.5x105N/C * 6m = -40.8J
 

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