A proton is released from rest in a uniform electric field of magnitude 8000 V/m

Question

A proton is released from rest in a uniform electric field of magnitude 8000 V/m directed along the positive x - axis. The proton undergoes a displacement of 0.5 m in the direction of the electric field as shown below. The charge on the proton is 1.602 X 10^-19 C, and its mass is 1.67 Times 10^-27 kg. Find the change in electric potential as the moves from point A to point B. Find the change in electric potential energy of the proton for this displacement. Find the work done by the electric field on the proton as it moves from point A to point B. Find the change in kinetic energy as the proton moves from point A to point B. Find the velocity of the portion at point B, assuming it starts from rest at point A.

Explanation / Answer

i)

VAB = E*d = 8000*0.5 = 4000 V

ii)

change in energy dU = dV*q = 4000*1.6*10^-19 = 6.4*10^-16 J

iii)

W = F*d = E*q*d = 8000*1.6*10^-19*0.5 = 6.4*10^-16 J

iV)

from work energy theorem

work = change in KE

dK = 6.4*10^-16 J

v)

initial kiknetic energy k1 = 0

final kinetic energy K2 = 0.5*m*v^2

0.5*m*v^2 = 6.4*10^-16

0.5*1.67*10^-27*v^2 = 6.4*10^-16

v = 8.75*10^5 m/s <<<<--------answer

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