Question

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.

Answer 1

Answer:

$A) \,K.E=1.405\times 10^{-20}J$

$B)\,r_f=0.268\,m$

Explanation:

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

A) Initial kinetic energy of proton

$K.E=\frac{1}{2}m_pv^2\\\\K.E=1.405\times 10^{-20}J$

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

Potential energy and kinetic energy are related as:

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

Change in voltage is

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