Given an Electric Field find the necessary Work to move a charge Q

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SUMMARY

The discussion centers on calculating the work required to move a charge Q=7μC in a specified electric field \(\vec{E}=z\vec{u_{x}}-3y^{2}\vec{u_{y}}+x\vec{u_{z}}\) V/m. The correct work calculation is derived using the formula \(W=-Q\int^{f}_{i}\vec{E}\cdot d\vec{l}\), resulting in a final answer of W=-75nJ. The error identified in the initial attempt was the failure to convert the differential path \(d\vec{l}\) into a unit vector, which led to an incorrect work value of W=-525nJ.

PREREQUISITES
  • Understanding of electric fields and their representation in vector form.
  • Familiarity with the concept of work done by a force in physics.
  • Knowledge of calculus, specifically integration of vector functions.
  • Ability to convert units, particularly between microcoulombs and coulombs.
NEXT STEPS
  • Review the principles of vector calculus as applied to electric fields.
  • Study the method of calculating work done in electric fields using integration.
  • Learn how to properly convert differential path elements into unit vectors.
  • Explore additional examples of work calculations in varying electric fields.
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering majors, and anyone involved in electromagnetism who seeks to understand the work-energy principle in electric fields.

kacete
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Homework Statement


Given the Electric field [tex]\vec{E}=z\vec{u_{x}}-3y^{2}\vec{u_{y}}+x\vec{u_{z}} V/m[/tex] find the necessary Work to move a charge [tex]Q=7\mu C[/tex] along an incremental path of [tex]1mm[/tex] distance in the direction of [tex]2\vec{u_{x}}-6\vec{u_{y}}-3\vec{u_{z}}[/tex] which locates in [tex]P(1,2,3)[/tex]

Homework Equations


Work
[tex]W=-Q\int^{f}_{i}\vec{E}.d\vec{l}[/tex]
i - inicial point
f - final point
Solution
[tex]W=-75nJ[/tex]

The Attempt at a Solution


Converting to base units
[tex]Q=7\mu C=7x10^{-6}C[/tex]
[tex]d=1mm=1x10^{-3}m[/tex]
The electric field at the given point
[tex]\vec{E}=3\vec{u_{x}}-12\vec{u_{y}}+1\vec{u_{z}} V/m[/tex]
The equation
[tex]W=-Q\int^{f}_{i}\vec{E}.d\vec{l}=-Q.E\int^{f}_{i}dl.(2\vec{u_{x}}-6\vec{u_{y}}-3\vec{u_{z}})[/tex]
where [tex]\int^{f}_{i}dl=1x10^{-3}[/tex]
My solution
[tex]W=-525nJ[/tex]

Where did I go wrong?
 
Physics news on Phys.org
You need to make dl a unit vector.
 
Can't believe it was that simple. I'm embarrassed. Thank you!
 

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