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

In summary, an electron initially at rest on the surface of a sphere with a radius of 1.0 cm and a uniformly distributed charge of 1.6x10^{-15} must have an escape speed of ##\sqrt{-2k\frac{q_1q_2}{mr}}## in order to have zero kinetic energy when it gets to an infinite distance from the sphere.
  • #1
Potatochip911
318
3

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).
 
Physics news on Phys.org
  • #2
You should know q1 and q2 by Gauss law.
 
  • #3
azizlwl said:
You should know q1 and q2 by Gauss law.
##q_1## is given and ##q_2=e##
 
  • #4
What is q1?
 
  • #5
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.
 
  • #6
There many charges distributed uniformly on the surface of the sphere.
 
  • #7
What could I possibly use surface charge density for in this question?
 
  • #8
Have you learn about Gauss's Law?
 
  • #9
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.
 

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

1. What is escape speed for an electron?

The escape speed for an electron is the minimum speed required for an electron to escape from a particular energy level or potential well. It is a measure of the energy needed for an electron to overcome the attractive force of the nucleus and move freely away.

2. How is escape speed for an electron calculated?

The escape speed for an electron can be calculated using the formula v = √(2E/m), where v is the escape speed, E is the initial energy of the electron, and m is the mass of the electron.

3. What factors affect the escape speed for an electron?

The escape speed for an electron is affected by the mass of the electron, the initial energy level, and the strength of the attractive force from the nucleus. Factors such as external electric or magnetic fields can also influence the escape speed.

4. Why is finding escape speed for an electron important?

Finding the escape speed for an electron is important because it helps us understand the behavior of electrons in different energy levels and potential wells. It also has practical applications in fields such as particle physics and astrophysics.

5. Can the escape speed for an electron be greater than the speed of light?

No, the escape speed for an electron cannot be greater than the speed of light. According to Einstein's theory of relativity, nothing can travel faster than the speed of light. Therefore, the maximum escape speed for an electron would be close to the speed of light, but not greater than it.

Similar threads

Back
Top