Calculate the proton’s initial kinetic energy.?

Question

1) An infinitely long line of charge has a linear charge density of (5.50 	imes 10^{-12} \, 	ext{C/m}). A proton, with a mass of (1.67 	imes 10^{-27} \, 	ext{kg}) and a charge of (1.60 	imes 10^{-19} \, 	ext{C}), is positioned 17.0 cm away from the line and is moving directly toward it at a velocity of 1000 m/s. 
a) Calculate the proton’s initial kinetic energy. 
b) Determine how close the proton can get to the line of charge.

Anonymous · Accepted Answer

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

Athena Beatty MD · Answer

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

Anonymous · Answer

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.

Anonymous · Answer

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Anonymous · Answer

________________________________________K.E of proton =0.5mv^2= 8.35*10^-22 JA) 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)drPotential difference = U2-U1 = integral -(2k*lembda)ln(r2/r1)____________________________________…Suppose proton stops at x m from line of chargeWork done by electric field = kinetic energywork =q(U2 -U1)=KE1.6*10^-19*(-2k*lembda)ln(r2/r1)=8.35*10^-… -(2k*lembda)ln(r2/r1)  =  8.35*10^-22 /1.6*10^-19ln(r2 / r1) = – 5.21875*10^-3 / 2k*lembdaln(r2/r1) = – 5.21875*10^-3/2k*lembdaln(r2/r1) = – 5.21875*10^-3/2*9*10^9*5.5*10^-12ln(r2/r1) = – 0.0527146ln(r2 /r1) = -0.0527146(r2 /r1) =   0.9486r2 = 0.17 *0.9486=0.16127mB) The proton gets up to 16.127 cm or 0.16127 m from the line of charge
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Anonymous

Calculate the proton’s initial kinetic energy.?

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