# Electrostatics charged particle Problems

1. Feb 26, 2005

### blue_soda025

1. A charged particle was accelerated from rest by a potential difference of 2.50 x 10^5 V. If the particle reached a maximum speed of 2.90 x 10^4 m/s, what potential difference would be required to accelerate this particle from rest to a velocity of 7.25 x 10^4 m/s?

2. The centers of two alpha particles are held 2.5 x 10^-12 m apart, then they are released. Calculate the speed of each alpha particle when they are 0.75 m apart.

For the second one, my teacher kind of went over it and came up with:
$$v = \sqrt{\frac{\frac{kq_{1}q_{2}}{r_{1}} - \frac{kq_{1}q_{2}}{r_{2}}}{m}}$$ (didn't want to show all the steps)
However, when I plug in the numbers, it doesn't match the answer which should be 2.4 x 10^5 m/s.

Last edited: Feb 26, 2005
2. Feb 27, 2005

### Davorak

1. Use energy conservation. Charge multiplied by potential difference gives an energy, this energy equals the kinetic energy. Use this equation and solve for charge times mass.
Now knowing the charge times mass for the particle find the potential difference that would be required to reach the second speed.

2. How much are you off by? Are you off by a factor of $\sqrt{2}$? Other then that it is hard to tell if anything else when wrong since you did not post the numbers you used.

3. Feb 27, 2005

### blue_soda025

Oops, I rechecked my numbers and I think some of the ones I entered were wrong. I redid it and I got the answer. Thanks for your help.