Charge Density of wire with potential difference

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Problem:

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 1.91 cm from the wire it has a speed 2.80E+5 m/s, going away from the wire. When it is at a new distance of 4.01 cm, its speed is 3.20E+6 m/s. What is the charge density of the wire?

Alright so this is how I approached it but didn't get it right:

The difference in kinetic energy at two given distances should be equal to the opposite difference in potential energy (\DeltaK = -\DeltaU), which when divided by the given charge of the particle should give the potential difference. Right? (\DeltaU = q*\DeltaV)

The potential difference is equal to -E*d where d is the distance between the two given points and E is the electric field of the wire given by: lambda/2*pi*epsilon*r.

With these you should be able to solve for lambda but I keep getting the answer wrong and I think its because I'm using the wrong r in the equation for the electric field. I thought r should be the distance from the wire to the farthest given point but I wasn't sure...

Can anyone help?
 
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Without showing your work, I would guess you may be finding the potential incorrectly. It looks like you are treating E as a constant with the expression V=-E*d. Recall,

V=-\int \mbox{Edr}

where r is the perpendicular distance from the wire. Your approach to the problem appears correct.
 
Yes I think I found my problem, I can't use the simplified V = -Ed equation because E is not a uniform electric field in this case.
 
Can you please explain it in more detail ? How do you calculate the potential difference then if you can't consider it a uniforn electric field ? and what is d ?
 
*unicorn* said:
Can you please explain it in more detail ? How do you calculate the potential difference then if you can't consider it a uniforn electric field ? and what is d ?
I'm guessing you don't know calculus. chrisk explained it. You must integrate the electric field as a function of distance. The "d" means "differential". If you don't know calculus, then I don't expect you will have any idea what this means.
 
ok.. nevermind.. i got it.. thanks !
 

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