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

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    ##U_0+K_0=U_f+K_f##
    ##U=k\frac{q_1q_2}{r}##

    3. 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).
     
  2. jcsd
  3. Dec 6, 2015 #2
    You should know q1 and q2 by Gauss law.
     
  4. Dec 6, 2015 #3
    ##q_1## is given and ##q_2=e##
     
  5. Dec 6, 2015 #4
    What is q1?
     
  6. Dec 6, 2015 #5
    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.
     
  7. Dec 6, 2015 #6
    There many charges distributed uniformly on the surface of the sphere.
     
  8. Dec 6, 2015 #7
    What could I possibly use surface charge density for in this question?
     
  9. Dec 6, 2015 #8
    Have you learn about Gauss's Law?
     
  10. Dec 7, 2015 #9
    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.
     
  11. Dec 7, 2015 #10
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