Force, Electric Field, and Positive Particle

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SUMMARY

The discussion focuses on calculating the location where an alpha particle experiences no net force due to a long line with a uniform linear charge density of 50.0 microcoulombs/m and a flat plastic sheet with a surface charge density of -100 microcoulombs/m². The relevant equations used include E = [1/(2*ε₀π)]*(λ/r) for the wire and E = ω/(2*ε₀) for the sheet. The solution derived indicates that the distance from the line where the alpha particle feels no force is approximately 0.1592 meters, positioned above the line due to the opposing directions of the electric field vectors.

PREREQUISITES
  • Understanding of electric fields generated by charged objects
  • Familiarity with linear charge density and surface charge density concepts
  • Knowledge of the equations for electric fields from infinite lines and sheets of charge
  • Basic calculus for solving equations involving electric fields
NEXT STEPS
  • Study the derivation of electric field equations for infinite charged lines and sheets
  • Learn about the concept of electric field superposition in electrostatics
  • Explore the behavior of charged particles in electric fields
  • Investigate the applications of electric fields in particle physics
USEFUL FOR

Students in physics, particularly those studying electromagnetism, as well as educators and anyone interested in understanding the interactions of charged particles in electric fields.

Soaring Crane
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Homework Statement



A long line carrying a uniform linear charge density 50.0 microcoulombs/m runs parallel to and 10.0 cm from the surface of a large, flat plastic sheet that has a uniform surface charge density of -100 microcoulombs/m^2 on one side.

Find the location of all points where an alpha particle would feel no force due to this arrangement of charged objects.
Indicate this location as a distance from the line.

Choose an appropriate location of these points at a distance: above the line or between the line and the sheet.

Homework Equations



E = [1/(2*episilon_0*pi)]*(lambda/r) for wire

E = omega/(2*episilon_0) for sheet (Do I use this formula?)


The Attempt at a Solution



I don’t know if I set this up correctly or used the correct formulas.

E = [1/(2*episilon_0*pi)]*(lambda/r) + omega/(2*episilon_0)

0 = [1/(2*episilon_0*pi)]*(lambda/r) + omega/(2*episilon_0)

-omega/(2*episilon_0) = [1/(2*episilon_0*pi)]*(lambda/r)

r = (lambda)/(-omega*pi) = (50*10^-6 C/m)/[pi -(-100*10^-6 C/m^2)] = 0.1592 m ?

This distance would be above the line because the electric field vectors, which are in same direction as force vectors, on the positive alpha particle would be in opposite directions??

Thanks.
 
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It looks fine to me. Is something unclear?
 

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