Answer
The electric force on the proton is in the +x direction.
Work Step by Step
We can find the electric field between $x = 1.0~cm$ and $x = 3.0~cm$:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{3.0~V-9.0~V}{0.030~m-0.010~m}$
$E = 300~V/m$
We can find the force on the proton:
$F = q~E$
$F = (1.6\times 10^{-19}~C)(300~V/m)$
$F = 4.80\times 10^{-17}~N$
The magnitude of the force on the proton is $~~4.80\times 10^{-17}~N$
The direction of the electric field is from higher potential to lower potential. In the region between $x = 1.0~cm$ and $x = 3.0~cm$, the electric field points in the +x direction.
Note that the charge on a proton is positive.
Therefore, the electric force on the proton is in the +x direction.
