Where Does an Electron Reach Equilibrium Between Two Charges?

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SUMMARY

The discussion focuses on calculating the equilibrium position of an electron between two point charges: -1.5 µC at the origin and 13 µC at coordinates (1 m, 0.5 m). The user attempts to find the x and y coordinates where the net force on the electron equals zero. The calculations reveal that the distance should be corrected to 1.118 m, leading to an x-coordinate of approximately 1.24e-7 m and a y-coordinate of 5.54e-8 m. The importance of clearly explaining the thought process behind calculations is emphasized for better understanding.

PREREQUISITES
  • Understanding of Coulomb's Law for calculating forces between charges
  • Familiarity with basic trigonometry for resolving forces into components
  • Knowledge of electric charge units, specifically microcoulombs (µC)
  • Ability to perform algebraic manipulations and solve equations
NEXT STEPS
  • Review Coulomb's Law and its application in electrostatics
  • Study the concept of electric field and potential due to point charges
  • Learn about vector resolution in physics for force calculations
  • Practice problems involving equilibrium of charged particles in electric fields
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics, as well as educators looking for examples of charge interactions and equilibrium calculations.

Zvaigzdute
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Homework Statement



A point charge of -1.5 mu or micro CC is located at the origin. A second point charge of 13 mu or micro CC is at x = 1 m, y = 0.5 m. Find the x and y coordinates of the position at which an electron would be in equilibrium.

Homework Equations





The Attempt at a Solution



Force acting on q2

F=k(-1.5e-6C)(13e-6C)/.118^2= -.140 F

Fnet= 0

.140=k(-1.5e-6)(-1.602177e-19)/x^2
x=1.24e-7

tan(x)=.5/1 x=26.56 degrees

sin(z) = y/1.24e-7

y=5.54 e -8

What am I doing wrong?
 
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Zvaigzdute said:

The Attempt at a Solution



Force acting on q2

F=k(-1.5e-6C)(13e-6C)/.118^2= -.140 F

Why? What do you need it for?
By the way, you're missing a "1" in the distance value. It should be 1.118m.

Fnet= 0

.140=k(-1.5e-6)(-1.602177e-19)/x^2
x=1.24e-7

tan(x)=.5/1 x=26.56 degrees

sin(z) = y/1.24e-7

y=5.54 e -8

What am I doing wrong?

Why don't you explain your intended steps? It's not obvious from the calculations you're performing what your thought process is.
 
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