Question

An infinitely long line of charge has a linear charge density of 5.50×10⁻¹² C/m . A proton is at distance 15.0 cm from the line and is moving directly toward the line with speed 1100 m/s .

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

Express your answer in meters.

Answer 1

at a distance x, electric field due to infinitely long line of charge=charge density/(2*pi*epsilon*x)

where epsilon=8.85*10^(-12) m/F

then electric field at a distance x=0.09891/x N/C

force on a proton at a distance x=electric field *charge

=0.09891*q/x

=0.09891*1.6*10^(-19)/x

=1.15826*10^(-20)/x

let denote the numerator as a constant A. (so that further calculations wont be too cumbersome to read)

then electric force=A/x and it is directed in opposite direction to the proton's movement.

as force=mass*v*dv/dx

==>m*v*dv/dx=A/x

==>v*dv=(A/m)*dx/x

A/m=1.15826*10^(-20)/(1.67*10^(-27))=6.9357*10^6

lets denote it by B

integrating both the sides,

v^2/2=B*ln(x)+c

where c is a constant of integration

if final distance is d, at that point v=0

hence using limits for v as 1100 m/s to 0 m/s and for x to be 0.15 m to d.

we get

(0^2-1100^2)/2=B*ln(d/0.15)

==>ln(d/0.15)=-0.08723

==>d/0.15=exp(-0.08723)=0.91647

==>d=0.15*0.91647=0.13747 m

hence the proton will stop at a distance of 0.13747 m from the line of charge.