An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C) is 12.0 cm from the line and moving directly toward the line at 4.10×103 m/s .

a)Calculate the proton's initial kinetic energy. Express your answer with the appropriate units.

b)How close does the proton get to the line of charge? Express your answer with the appropriate units.

[tex]Charge\,\,density=\lambda=6\times 10^{-12}C/n\\\\Mass\,\, of \,\,proton=m_p=1.67\times 10^{-27}kg\\\\charge\,\, of\,\, proton=q_p=1.609\times 10^{-19}C\\\\r=12\,cm=0.12\, m\\\\v=4.103\times 10^{3}m/s[/tex]

A) Initial kinetic energy of proton

[tex]K.E=\frac{1}{2}m_pv^2\\\\K.E=1.405\times 10^{-20}J[/tex]

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

Potential energy and kinetic energy are related as:

[tex]K_i+U_i=K_f+U_f\\\\U_f-U_i=K_i-K_f\\\\q(V_f-V_i)=1.40\times 10^{-20}\\\\V_f-V_i=0.087--(1)\\[/tex]

Change in voltage is

[tex]V_f-V_i=\frac{\lambda}{2\pi \epsilon_o}ln\frac{r_f}{r_i}\\\\ln|\frac{r_f}{r_i}|=(0.087)(\frac{2\pi \epsilon_o}{\lambda})\\\\ln|\frac{r_f}{r_i}|==0.8059\\\\\frac{r_f}{r_i}=2.24\\\\r_f=(2.24)(.12)\\\\r_f=0.268 m[/tex]

ConcepcionPetillo ConcepcionPetillo · Answer 2 · 2022-04-08T15:22:27+02:00

As the proton approaches the line of charge:

(a) The initial kinetic energy of the proton is 1.4×10⁻²⁰J

(b) Distance of closest approach is 0.266m

Calculating the distance of the closest approach:

Given that the linear charge density of the line is λ = 6×10⁻¹² C/m

mass of the proton, m = 1.67×10⁻²⁷kg

charge on proton, q = 1.6×10⁻¹⁹ C

initial distance of the proton, r = 12cm = 0.12m

velocity of the proton, v = 4.1×10³ m/s

(a) Initial kinetic energy of the proton is given by:

KE = ¹/₂mv²

KE = 0.5×(1.67×10⁻²⁷)×(4.1×10³)²

KE = 1.4×10⁻²⁰J

(b) When the proton is closest to the line of charge, its velocity is zero, so according to the conservation of energy, all the kinetic energy is converted into potential energy

The change in potential energy due to linear charge density is given by:

[tex]\Delta U=\frac{\lambda q}{4\pi\epsilon_0}\ln\frac{r'}{r}[/tex]

where r' is the final position

since ΔU = KE = 1.4×10⁻²⁰J

[tex]1.4\times10^{-20}=9\times10^9\times6\times10^{-12}\times1.67\times10^{-27}\ln\frac{r'}{0.12}\\\\\ln\frac{r'}{0.12}=0.8\\\\\frac{r'}{0.12}=2.22[/tex]

r' = 0.266m

So, the proton gets as close as 0.266m to the line of charge.

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