An infinitely long line of charge has linear charge density 5.00 10-12 C/m. A proton (mass 1.67 10-27 kg, charge +1.60 10-19 C) is 18.0 cm from the line and moving directly toward the line at 1.20 103 m/s. How close does the proton get to the line of charge?

16.57cm

Explanation:

Given

λ = 5*10^-12C/m

Mass, m = 1.67*10^-27kg

v = 1.2*10^3

Ri = 18cm

K = ?

Rf = ?

K = mv²/2

K = (1.67*10^-27)*(1.2*10^3)²/2

K = 1.2*10^-21J

Recall, Vf - Vi = K/e

Vf - Vi = 1.2*10^-21/1.6*10^-19

Vf - Vi = 7.5*10^-3

Vf - Vi = 0.0075

Also, Vf - Vi = (λ/2πE)[In(Rf/Ri)]

In Rf/Ri = (-0.0075*2*π*8.85*10^-12)/5*10^-12

In Rf/Ri = -0.083

Rf = Ri (exp -0.083)

Rf = 18 (exp -0.083)

Rf = 16.57 cm

the proton gets as close as 16.57cm to the line of charge

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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