# What is the kinetic energy of the proton?

## Homework Statement

An infinitely long line of charge has linear charge density λ = 3.00 pC/m. A proton is at
a distance d = 14.5 cm from the line and moving directly toward the line at v = 2.00 km/s.
a. (3 points) What is the kinetic energy of the proton?
b. (15 points) How close does the proton get to the line of charge?

## Homework Equations

K=.5mv2
E(line of charge)=($$\lambda$$/(2$$\Pi$$$$\epsilon$$subnought))*1/r
W=Kfinal-Kinitial
W=Uinitial-Ufinal
U=k(q1q2)/r

## The Attempt at a Solution

First let me apologize for my lack of understanding how to input formulas into the forum. I found the latex reference tool thing but when I click something it just inserts a bunch of garbletygoop that makes my head explode a little. Yes, garbletygoop.

Part a is just a simple application of the kinetic energy equation, yielding 3.34×10^−21 J. This part is just a lead in for how to work part b. To find out how close the proton comes to the line of charge, I realized that we are really looking for the turning point of the proton, or the point at which kinetic energy equals zero. Since W=Kfinal-Kinitial, this means the work done is -Kinitial, which we just solved for in part a.

Next I plugged this into the equation W=Uinitial-Ufinal. So, -Kinitial=Uinitial-Ufinal. I know the distance (r) for the Uinitial equation, and the distance (r) for the Ufinal equation is my variable. Seemed pretty straight forward. But then I realized that my electric potential energy equations were for two point charges, not for a point charge and a line of charge. I can find no such equation, and honestly my calculus skills are not such that I can derive one of my own. Suggestions?