Electron Approaching a Charged Sphere

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SUMMARY

The discussion focuses on calculating the final velocity of an electron as it approaches a uniformly charged sphere with a total charge of 1.01×10-9 C. The electron starts from rest at a distance of 2.80 cm from the center of the sphere, which has a radius of 1.83 cm. Using the conservation of energy principle, the final velocity is determined to be approximately 7.9×107 m/s. The calculations involve the constants k = 9×109 N·m²/C², e = 1.602×10-19 C, and m = 9.1094×10-31 kg.

PREREQUISITES
  • Understanding of electrostatic potential energy and kinetic energy
  • Familiarity with the conservation of energy principle
  • Knowledge of Coulomb's law and electric fields
  • Basic algebra for rearranging equations
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  • Learn about electric potential due to point charges and spherical charge distributions
  • Explore the implications of electric fields on charged particles
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This discussion is beneficial for physics students, educators, and anyone interested in electrostatics and particle dynamics, particularly in understanding the behavior of charged particles in electric fields.

jrk012
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Homework Statement



An electron starts from rest 2.80 cm from the center of a uniformly charged sphere of radius 1.83 cm. If the sphere carries a total charge of 1.01×10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?


Homework Equations



UI + KI = UF + KF
U=(k*Q*-e)/r
K=1/2mv^2


The Attempt at a Solution



Rearranging terms I got:

[([(k*Q*-e)/(r-initial)] - [(k*Q*-e)/(radius)])/(r-initial - radius)](2/m)

Using:
k = 9x10^9
Q = 1.01×10-9
e = 1.602 x 10^(-19)
m = 9.1094 x 10^(-31)
r-initial = .028m
radius = .0183m

I then get v^2 = 6.24x10^15. I square root(√) it to get v and I get:

v = 7.9x10^7 m/s

I have tried the negative and positive of this and I cannot solve it. Any help would be appreciated
 
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jrk012 said:

Homework Statement



An electron starts from rest 2.80 cm from the center of a uniformly charged sphere of radius 1.83 cm. If the sphere carries a total charge of 1.01×10-9 C, how fast will the electron be moving when it reaches the surface of the sphere?


Homework Equations



UI + KI = UF + KF
U=(k*Q*-e)/r
K=1/2mv^2


The Attempt at a Solution



Rearranging terms I got:

[([(k*Q*-e)/(r-initial)] - [(k*Q*-e)/(radius)])/(r-initial - radius)](2/m)

Using:
k = 9x10^9
Q = 1.01×10-9
e = 1.602 x 10^(-19)
m = 9.1094 x 10^(-31)
r-initial = .028m
radius = .0183m

I then get v^2 = 6.24x10^15. I square root(√) it to get v and I get:

v = 7.9x10^7 m/s

I have tried the negative and positive of this and I cannot solve it. Any help would be appreciated
What is the factor, (r-initial - radius) doing in there?
 
SammyS, you are a saint. Thank you!
 

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