Question

"Consider a long charged straight wire that lies fixed and a particle of charge +2e and mass 6.70×10-27 kg. When the particle is at a distance 1.61 cm from the wire it has a speed 2.70×105 m/s, going away from the wire. When it is at a new distance of 2.71 cm, its speed is 3.40×106 m/s. What is the charge density of the wire?"

This is my attempt.

From conservation of energy we can find the electric field:

KE(i) + U(i) = KE(f) + U(f)

1/2mv(i)2 + qV(i) = 1/2mv(f)2 + qV(f) (v here is speed, V is potential)

Some algebra leads us to this:

Δ(V) = m(v(i)2 - v(f)2 )/(2q)

Since E = Δ(V)/Δ(d)

Then:

E = m(v(i)2 - v(f)2 )/(2q * Δ(d))

I find the same answer using kinematics:

F = ma = qE and v(f)2 = v(i)2 + 2aΔd so we can find figure out E

Finally, using Gauss's law:

The field generated by an infinite (long) wire is:

E = lambda/(2piR*epsilon)

Plugging in my expression for E and working out the algebra, I end up with:

lambda = m(v(i)2 - m(f)2 )*2piR(epsilon)/(2qΔd)

Which "R" do I use here? In other words, what is the radius of my Gaussian surface? Or is my entire approach completely wrong?