What is the Escape Speed of an Electron from a Charged Glass Sphere?

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SUMMARY

The escape speed of an electron from a charged glass sphere with a diameter of 7.42 cm and a charge of 7.04 nC can be calculated using the formula v = √(2kqq/mr). The correct values for the charge and diameter must be used, with the mass of the electron being 9.11 x 10^-31 kg. The user initially interchanged the charge and diameter values, leading to incorrect calculations. The correct radius for the calculation is 0.0371 m.

PREREQUISITES
  • Understanding of electrostatics, specifically Coulomb's law.
  • Familiarity with kinetic energy equations in physics.
  • Knowledge of the mass of an electron (9.11 x 10^-31 kg).
  • Ability to manipulate algebraic equations to solve for variables.
NEXT STEPS
  • Review Coulomb's law and its application in electrostatics.
  • Study the derivation and application of kinetic energy equations.
  • Practice problems involving escape velocity calculations for charged objects.
  • Explore the implications of charge distribution on electric fields and potential energy.
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics and energy conservation, as well as educators looking for examples of escape velocity calculations involving charged particles.

kmikias
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Homework Statement



What is the escape speed (in m/s) of an electron launched from the surface of a 7.42 cm diameter glass sphere that has been charged to 7.04 nC?

Homework Equations



u=kqq/r
KE=1/2MV^2
UO=KE

The Attempt at a Solution


here is what i did
kqq/r = 1/2mv^2
solve for v
v=\sqrt{}2kqq/rm
i used q = 7.42 square
and mass of electron = 9.11*10^-31
and r = 0.0704/2
...but still my answer is wrong
 
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kmikias said:
What is the escape speed (in m/s) of an electron launched from the surface of a 7.42 cm diameter glass sphere that has been charged to 7.04 nC?

kmikias said:

The Attempt at a Solution


here is what i did
kqq/r = 1/2mv^2
solve for v
v=\sqrt{}2kqq/rm
i used q = 7.42 square
and mass of electron = 9.11*10^-31
and r = 0.0704/2
...but still my answer is wrong

I believe you have the value of the sphere's charge, and the diameter of the sphere interchanged.
 

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