Finding Escape Speed for an Electron that's Initially at Res

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Homework Help Overview

The problem involves determining the escape speed for an electron that is initially at rest on the surface of a charged sphere with a specified radius and charge. The context is rooted in electrostatics and energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy and kinetic energy, questioning the appropriate use of the radius in the potential energy formula. There is also confusion regarding the initial kinetic energy of the electron when it is stated to be at rest.

Discussion Status

Participants are exploring the definitions of the charges involved and the application of Gauss's Law. Some have provided guidance on identifying the charges, while others are questioning the relevance of surface charge density in this context. Multiple interpretations of the problem setup are being examined.

Contextual Notes

There is a discussion about the implications of the radius in the potential energy equation, with some participants noting that the electron is at the surface of the sphere, which raises questions about the separation distance. The conversation also touches on the uniform distribution of charge on the sphere and its effect on the electric field.

Potatochip911
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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 distributed charge of ##1.6\times 10^{-15}##? 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


##U_0+K_0=U_f+K_f##
##U=k\frac{q_1q_2}{r}##

The Attempt at a Solution


$$U_0+K_0=U_f+K_f\Longrightarrow U_0+K_0=0+0\Longrightarrow K_0=-U_0 \\
\frac{1}{2}mv^2=-k\frac{q_1q_2}{r} \\
v=\sqrt{-2k\frac{q_1q_2}{mr}}$$
This is the correct answer but I'm confused a bit as to the radius ##r##, in my textbook it states that in the formula for potential energy, ##U=k\frac{q_1q_2}{r}## that ##r## is the separation between the two particles, however in the answer they plug in ##r=0.01m##, but I would think that the separation between the electron and the sphere is ##0## since the electron is initially at rest on the surface of the sphere (I also don't understand how it can have initial kinetic energy if it's at rest on the sphere).
 
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You should know q1 and q2 by Gauss law.
 
azizlwl said:
You should know q1 and q2 by Gauss law.
##q_1## is given and ##q_2=e##
 
What is q1?
 
Potatochip911 said:
surface of a sphere with a radius of 1.0 cm and a uniformly distributed charge of ##1.6\times 10^{-15}##
Personally I denoted the charge of the sphere as ##q_1=1.6\times 10^{-15}## and ##q_2=e## although since we are dealing with only two charges it doesn't matter which one is labelled which.
 
There many charges distributed uniformly on the surface of the sphere.
 
What could I possibly use surface charge density for in this question?
 
Have you learn about Gauss's Law?
 
azizlwl said:
Have you learn about Gauss's Law?
Yes for a sphere $$\oint \vec{E}\cdot\vec{dA}=\frac{q_{encl}}{\varepsilon_0} \\
E(4\pi r^2)=\frac{q_{encl}}{\varepsilon_0} $$ I just don't see where this is going.
 

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