A proton is released from rest at #x = -2 cm# in a constant electric field with magnitude #1500# #N##/C# pointing in the positive #x# direction. What is the speed of the proton at #x= 5cm#?

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Answer 1 · 2016-10-31T17:18:15

#approx 142 # km/s

From Newton's Law

#vec F = m_p vec a = q_p vecE implies a = q_p/m_p E# in the positive x-direction

and we can use the equation of motion:

#v^2 = u^2 + 2 a Deltax#

...with #Delta x = +0.07# in the positive x-direction, and #u = 0#

Putting it all together:

#v = sqrt(2 a Delta x) #
#= sqrt (2 * (1.6022 x 10^(-19))/(1.6726 x 10^(-27)) * 1500 * 0.07 )#

#approx 142 # km/s

This is still way off relativistic so I think we can leave it there.