Gauss Sphere Surface / U = -W / f=ma=qE

AI Thread Summary
The discussion centers on calculating the escape speed for an electron on the surface of a charged sphere with a radius of 1.0 cm and a charge of 1.6 X 10^-15 C. Participants explore the relationship between electric potential energy and kinetic energy, emphasizing that the electron must reach a point of zero kinetic energy at infinity. Clarification is sought on why the problem states that the electron has no kinetic energy at infinity, leading to the understanding that this refers to the minimum speed required to escape the sphere's gravitational influence. The equations governing potential energy and work are discussed, with participants sharing resources to aid in comprehension. Overall, the conversation highlights the complexities of energy conservation in electrostatics and the importance of understanding these principles in physics.
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Homework Statement


What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.0 cm and a uniformly distrubted charge of 1.6 X 10^-15 C? That is, what initial speed must the electron have in order to reach an infinite distance from the sphere and have zero kinetic energy when it gets there?


Homework Equations


E = spherical surface with q charge: 1/2Eo (q/r)
U = -W
eV = Winfinity->A
Ke = 1/2(mv^2)
f=ma=qE
Vf + Vi + Ui + Uf = 0

The Attempt at a Solution


I've played with all these relevant equations and tried to massage the numbers, but I'm pretty off.

I know that we have equilibrium while the electron is on the charged sphere surface. We have Vi. When we reach point infinite distance from sphere we have Vf...why does it have zero kinetic energy when it gets there if Kinetic energy is different from Work? Doesn't that mean it has no mass and no velocity?

The last relevant equation gives me ideas that I should be working in two parallel equations.

Thanks in advance. It's really great to read through all the forums in here and see some of the fun challenges that lie ahead in Physics - harmonic oscillations, divergence/convergence.
 
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potential energy

What electric potential energy does the electron have at the surface of the sphere? (How does potential energy depend on distance?)
 
Ui = 0 at infinity and electric potential V is also zero at infinity.
Uf/q = electric potential energy on the surface of the sphere.
 
Not sure what you're saying here. Initially, at the surface of the sphere, what's the potential energy of the electron? The final potential energy at infinity will be zero.

You might want to read this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html"
 
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Thank you for the link; that is clear and helpful (and unlike anything in my book's chapter).

I was trying to say similarly I think, applying Gauss/Coulomb ideas with:

Uf - Ui = -W

I'll try writing the math now!

Question though, why does the problem statement say that there is no "kinetic" energy when the electron reaches the infinite distance? I just saw something about this in one of the threads in here...about the electron no longer having mass at this infinite point. 0 Kinetic energy would be achieved also if there is no final velocity.
 
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escape speed is the minimum needed to reach infinity

mewon said:
Question though, why does the problem statement say that there is no "kinetic" energy when the electron reaches the infinite distance?
Because they want you to calculate the minimum speed (or kinetic energy) needed to escape and reach "infinity". That's the speed for which it just runs out of energy as it goes to infinity. You can always give the electron a greater initial speed, but then it will have more than enough energy and will never slow down to zero speed or kinetic energy.
 
Thank you Doc Al!

Had to take a break but I got it now ;-) It was easier than I was making it.

U = W
1/2(MV^2) = (1/4pieEo)(Qe/r)
 
This is a great site.

I'm paying my dues which are a great value and I'm sure I'll be here more through the next quarter Physics curriculum as well.

I am teaching post MA and have trouble getting to my Physics teacher office hours nor any school offered tutoring. It's also a way better deal and more convenient than hiring a local tutor to help on any HW problems I have questions on!
 
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