Question

An infinitely long line of charge has linear charge density λ = 2 pC/m. A proton is at a distance d = 14.5 cm from the line and moving directly toward the line at v = 5.3 km/s. The mass and charge of a proton are 1.67E-27 kg and 1.602E-19 C, respectively.

When entering in your answers, use the notation 1.3E-67 to represent 1.3 x 10^-67, for example.

What is the kinetic energy of the proton in Joules?

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

Answer 1

Answer:

Given, linear charge density λ = 2 pC/, initial distance between the line and the proton is d₁ = 14.5 cm = 0.0145 m and velocity of the proton is v = 5.3 km/s = 5300 m/s, mass of the proton m = 1.67 x 10^-27 kg and charge of a proton q = 1.602 x 10^-19 C

(a) The kinetic energy of the proton is

K₁ = 1/2 mv² = 1/2 (1.67 x 10^-27 kg) (5300 m/s)² = 2.34 x 10^-20 J = 2.34E^-20 J

(b) Using the conservation of energy,

E_initial = E_final

K₁ + U₁ = K₂ + U₂

The final kinetic energy of the proton is zero.

Thus, U₂ - U₁ = K₁

e (V₂ - V₁) = K₁

or electric potential V₂ - V₁ = K₁/e ---------- (1)

The electric potential due to an infinite line charge at a point 1 with respect to point 2 is

V₁ - V₂ = (λ/2$\pi \varepsilon$₀) ln(d₂/d₁) -------------------- (2)

Eq.(1) becomes, V₁ - V₂ = -K₁/e

Thus, (λ/2$\pi \varepsilon$₀) ln(d₂/d₁) = -K₁/e

ln(d₂/d₁) = -2$\pi \varepsilon$₀K₁ / λe = -2$\pi$(8.85 x 10^-12 C²/N.m²) (2.34 x 10^-20 J) / (2 x 10^-12 C/m) (1.6 x 10^-19 C)

ln(d₂/d₁) = -4.064

or d₂/d₁ = e^-4.064

or d₂ = d₁[ e^-4.064 ]

Therefore, d₂ = (14.5 cm) e^-4.064 = 0.249 cm = 2.49 x 10^-3 m = 2.49E^-3 m

Hence, this is the closest distance between the line charge and the proton.