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## Homework Statement

A proton enters the uniform electric field produced by the two charged plates shown below. The magnitude of the electric field is 4.0 × 105 N/C, and the speed of the proton when it enters is 1.5 × 107 m/s. What distance d has the proton been deflected downward when it leaves the plates?

## Homework Equations

E = F/q

F = ma

PE = mgh

KE = 1/2mv^2

## The Attempt at a Solution

*mgh = 1/2mv^2*

gh = 1/2v^2

h = v^2 / 2g

Original height: h = (1.5*10^7)^2 / 2(9.8) = 1.15*10^13 m

E = F/q = ma/q

Eq / m = a

a = [(4*10^5) * (-1.6*10^-19)] / (1.673*10^-27)

a = -3.83 m/s^2

gh = 1/2v^2

h = v^2 / 2g

Original height: h = (1.5*10^7)^2 / 2(9.8) = 1.15*10^13 m

E = F/q = ma/q

Eq / m = a

a = [(4*10^5) * (-1.6*10^-19)] / (1.673*10^-27)

a = -3.83 m/s^2

I'm not really sure where to go from here. Would I need time to get velocity from the acceleration I found? I'm just trying to find velocity for the proton with the force of the electric field acting on it and then using that in the height equation (which I'm not sure is the right one to use here) to find the difference between the original height where the proton goes straight forward (no electric field).

I also know the force of an electric field and can use its velocity to determine the time that the proton takes to travel through the 12 cm (or 12 * 10^-6 m) space but I have no idea how to compute the vertical distance it takes depending on the downward force exerted by the field.

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