Question

A uniform electric field has a magnitude of 2.19E+3 N/C. In a vacuum, a proton begins with a speed of 2.74E+4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 2.10 mm.

Answer 1

As proton moves (attracted) along the E field. Work will be done by the field which will result in increase in the speed of proton. As the distance reduces the force gets stronger.

work done by electric force field = gain in KE of proton
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basically conservation of energy in conservative force field
PE + KE = const
d(PE) + d(KE)=0
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Force * displacement = KE(final) - KE(initial)
q E * 2.10*10^-3 = 0.5 m[vf^2 - vi^2]
[vf^2 - vi^2] = 2 q E * 2.10*10^-3 /m
[vf^2-vi^2]=2*1.6*10^-19*2.19*10^3*2.10*10^-3/1.67*10^-27
[vf^2-vi^2]=8.812*10^8
vf^2 = 8.812*10^8 + ( 2.74E+4)^2= 16.31*10^8
vf = 4.0385*10^4 m/s