Electric Potential Homework - Alpha Particle & Electron Impact Speeds

[raindrops]

Homework Statement

A 2mm diameter plastic bead is charged to -1nC (*note - the bead is not hollow)

a) An alpha particle is fired at the bead from far away with a speed of 1x10⁶ m/s and it collides head-on. What is the impact speed?

b) An electron is fired at the bead from far away. It "reflects", with a turning point .1mm from the surface of the bead. What was the electrons initial speed?

Homework Equations

V=PE_e/q V=kq/r W=k q1 q2 (1/ri-1/rf) F=ma E=F/q E=k[q]/r²

I don't know which of these equations are even needed, but I thought I would include as many relevant ones I could think of.

The Attempt at a Solution

I don't even know where to start with this. The only thing I can think of would be to maybe do something with conservation of energy? but I am given no distances for anything so I'm not sure how to make that or any of the above equations work.

 

[anathema]

Hello,

Thank you for your question. This is a very interesting problem and it requires some understanding of electrostatics and mechanics. Let's break it down step by step.

a) An alpha particle is fired at the bead from far away with a speed of 1x106 m/s and it collides head-on. What is the impact speed?

To solve this problem, we need to use the principle of conservation of energy. The initial energy of the alpha particle is given by its kinetic energy, which is 1/2 mv^2. We can calculate the mass of the alpha particle using its atomic weight and the speed given. The final energy of the alpha particle after the collision is given by its kinetic energy plus its potential energy due to the electric field created by the charged bead. The potential energy can be calculated using the equation V=kq/r, where k is the Coulomb constant, q is the charge on the bead, and r is the distance between the bead and the alpha particle. Since the alpha particle is colliding head-on with the bead, the distance r is equal to the diameter of the bead, which is 2mm. We can now set the initial energy equal to the final energy and solve for the impact speed.

b) An electron is fired at the bead from far away. It "reflects", with a turning point .1mm from the surface of the bead. What was the electrons initial speed?

This problem requires a deeper understanding of mechanics and electrostatics. When an electron is fired at the bead, it experiences a force due to the electric field created by the charged bead. This force causes the electron to change its direction and eventually "reflect" back. The turning point of the electron is when the force due to the electric field is equal to the force due to the initial velocity of the electron. We can use the equation F=ma to calculate the force due to the initial velocity of the electron. We can also use the equation E=F/q to calculate the electric field at the turning point. Once we have the electric field, we can use the equation V=PEe/q to calculate the potential energy at the turning point. The potential energy at the turning point is equal to the initial kinetic energy of the electron. We can now set the initial kinetic energy equal to the potential energy at the turning point and solve for the initial speed of the electron.

I hope this helps you solve the problem. Let me