Question

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 e) is 14.5 cm from the line and moving directly toward the line at 2500 m/s .

Calculate the proton’s initial kinetic energy.

How close does the proton get to the line of charge?

Please show how you got the answer so that I understand. :)

Answer 1

Given that :

linear charge density, = 5 x 10^-12 C/m

mass of proton, m_p = 1.67 x 10^-27 kg

velocity, v = 2500 m/s

distance of proton, d = 14.5 cm = 0.145 m

(a) The proton’s initial kinetic energy which will be given as :

K.E = (1/2) m_p v²                                                                       { eq.1 }

inserting the values in above eq.

K.E = (0.5) (1.67 x 10^-27 kg) (2500 m/s)²

K.E = 5218750 x 10^-27 J

K.E = 5.21 x 10^-21 J

(b) The proton get to the line of charge at closest distance which is given as :

using a formula, r = d e^{-(mv^2/ 4kq)}                                                                      { eq.2 }

where, q = charge on proton = 1.6 x 10^-19 C

k = constant = 9 x 10⁹ Nm²/C²

inserting the values in eq.2,

r = (0.145 m) e^{-{(1.67 x 10^-27 kg) x (2500 m/s)^2 / 4 (9 x 10^9 Nm^2/C^2) (5 x 10^-12 C/m) (1.6 x 10^-19 C)}}

r = (0.145 m) e^{-(3.62 x 10^-4)}

r = (0.145 m) (0.9996)

r = 0.145 m