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

1. Dec 6, 2015

Potatochip911

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. Dec 6, 2015

azizlwl

You should know q1 and q2 by Gauss law.

3. Dec 6, 2015

Potatochip911

$q_1$ is given and $q_2=e$

4. Dec 6, 2015

What is q1?

5. Dec 6, 2015

Potatochip911

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.

6. Dec 6, 2015

azizlwl

There many charges distributed uniformly on the surface of the sphere.

7. Dec 6, 2015

Potatochip911

What could I possibly use surface charge density for in this question?

8. Dec 6, 2015

azizlwl

Have you learn about Gauss's Law?

9. Dec 7, 2015

Potatochip911

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.

10. Dec 7, 2015