# Finding charge density of a wire

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!

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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).

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.