An infinitely long line of charge has linear charge density 4.50×10−12 Cm. A proton (mass 1.67×10−27 , charge 1.60×10−19 C) is 16.5cm from the line and moving directly toward the line at 1000m/s .
A) Calculate the proton’s initial kinetic energy.
B) How close does the proton get to the line of charge?
2 Answers
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E=2k*lembda/r
Potential difference = dU = -Edr= -(2k*lembda/r)dr
Potential difference = U = -(2k*lembda)ln(r)
K.E of proton =0.5mv^2= 8.35*10^-22 J
A) The proton’s initial kinetic energy is 8.35*10^-22 J
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Suppose proton stops at x m from line of charge
Work done by electric field = kinetic energy
work =qU=KE
1.6*10^-19*(-2k*lembda)ln(r)=8.35*10^-22
-(2k*lembda)ln(r)=8.35*10^-22 /1.6*10^-19
ln(r1) - ln(r2) = - 5.21875*10^-3 / 2k*lembda
ln(r1) - ln(r2) = - 5.21875*10^-3/2k*lembda
ln(r1) - ln(r2) = - 5.21875*10^-3/2*9*10^9*4.5*10^-12
ln(r1) - ln(r2) = -0.064429
ln(r1 /r2) = -0.064429
(r1 /r2) =0.9376
r1 = 0.165 *0.9376=0.1547m
B) The proton gets up to 15.47 cm or 0.1547 m from the line of charge
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