# Finding charge density of a wire

hanagasumi
1. Consider a long charged straight wire that lies fixed and a particle of charge +2e and mass 6.70E-27 kg. When the particle is at a distance 1.61 cm from the wire it has a speed 2.20E+5 m/s, going away from the wire. When it is at a new distance of 4.01 cm, its speed is 4.00E+6 m/s.

2. What is the charge density of the wire?

3. I attempted to do this question but had no idea how would I approach this, I thought:
1) I find total KE by using 1/2(mv^2+mv'^2), and this KE is equivalent to the -U of the system,
2)I know that dU=Q*dV
3) dV= - (integral) E*ds, but I'm not sure which distance I should use.
4)I know that E=lambda/(2pi*r*e0)

Then I have no idea how should I interconnect my information, and I'm confused if about the change in KE and U, and the actualy KE and U. Also, if we are supposed to integrate this, I don't know how to define the upper and lower limit for the function.

Thank you very much in advance!

JesseC
In this case the potential energy lost by the charge will be equal to its change in kinetic energy.

You know the electric field due to a long straight wire, it might help you to find the electrostatic potential V. You've written down how it relates to E, it means integrating the E field along a line.

What do you think would be an appropriate line to integrate along? Think about how the charge is moving away from the wire! (It starts a some distance 'a' away from the wire, then is found again a new distance 'b' away).

hanagasumi
In this case the potential energy lost by the charge will be equal to its change in kinetic energy.

You know the electric field due to a long straight wire, it might help you to find the electrostatic potential V. You've written down how it relates to E, it means integrating the E field along a line.

What do you think would be an appropriate line to integrate along? Think about how the charge is moving away from the wire! (It starts a some distance 'a' away from the wire, then is found again a new distance 'b' away).

Thank you so much! I'm kind of getting this now, I'll go try it again.