A proton is released in a uniform electric field between two parallel plates of

Question

A proton is released in a uniform electric field between two parallel plates of opposite charge. If the magnitude of the uniform electric field is 8.0 * 10^4 V/m, directed along the positive x axis, then the proton undergoes a displacement of 0.50 m in the direction of the field.
(a.) Find the change in electric potential of the proton as a result of this displacement.
(b.) Find the change in potential energy of the proton for this displacement.
(c.) Find the speed of the proton after it has moved 0.50 m, starting from rest.

Explanation / Answer

a) = V = Ed = (8*10^4 V/m)(0.5m) = 4*10^4 V

b) U = qV = (1.602*10^-19C)(4*10^4V) = 6.4*10^-15 J

c) Your potential energy is actually negative (I know this seems weird, because it doesn't follow the formula, but think about it this way: your potential energy initially was zero, and now it's some value; that value NEEDS to be negative because potential energy ALWAYS decreases for spontaneous processes: the proton WANTS to go down the potential gradient), so if you drop in potential energy, you need to gain kinetic energy:

K = (1/2)mv^2 = 6.4*10^-15 J

So, v = (2*(6.4*10^-15 J)/(1.67*10^-27kg)) = 2.77*10^6 m/s

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