Question

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`?

Answer 1

`approx 142` km/s

Explanation:

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.