1) An infinitely long line of charge has linear charge density 5.50×10−12 C/m. A proton (mass 1.67×10−27 kg , charge 1.60×10−19 C ) is 17.0 cm from the line and moving directly toward the line at 1000 m/s. a) Calculate the proton’s initial kinetic energy. b)How close does the proton get to the line of charge?

### 5 Answers

A) Protons mass: 1.67 x 10^-27kg V= 1000 m/s K.E= 1/2*m*v^2 =0.5*1.67*10^-27kg*(1000)^2 = 8.35*10^-22 J A) The proton’s initial kinetic energy is 8.35*10^-22 J

kinetic energy = 0.5mv^2 K.E= 0.5* 91.67*10^-27 ) *100^2

For a given velocity, maximum range R = v^2 / g and for a given range this is the minimum velocity => Minimum velocity, v = √(Rg) = √(98*9.8) = 31 m/s => Minimum KE = (1/2)*(0.8)*(31) J = 12.4 J Average force = 12.4/2 = 6.2 N You can use this method to work out the other two cases.

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________________________________________ 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 _________________________——– Electric field due to a line of charge = E=2k*lembda/r dU = -Edr= -(2k*lembda/r)dr Potential difference = U2-U1 = integral -(2k*lembda)ln(r2/r1) ____________________________________… Suppose proton stops at x m from line of charge Work done by electric field = kinetic energy work =q(U2 -U1)=KE 1.6*10^-19*(-2k*lembda)ln(r2/r1)=8.35*10^-… -(2k*lembda)ln(r2/r1) = 8.35*10^-22 /1.6*10^-19 ln(r2 / r1) = – 5.21875*10^-3 / 2k*lembda ln(r2/r1) = – 5.21875*10^-3/2k*lembda ln(r2/r1) = – 5.21875*10^-3/2*9*10^9*5.5*10^-12 ln(r2/r1) = – 0.0527146 ln(r2 /r1) = -0.0527146 (r2 /r1) = 0.9486 r2 = 0.17 *0.9486=0.16127m B) The proton gets up to 16.127 cm or 0.16127 m from the line of charge

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