Answer
The magnitude of the force on the proton is $~~3.20\times 10^{-17}~N$
Work Step by Step
We can find the strength of the electric field between $x = 5.0~cm$ and $x = 6.0~cm$:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{5.0~V-3.0~V}{0.060~m-0.050~m}$
$E = -200~V/m$
We can find the force on the proton:
$F = q~E$
$F = (1.6\times 10^{-19}~C)(-200~V/m)$
$F = -3.20\times 10^{-17}~N$
The magnitude of the force on the proton is $~~3.20\times 10^{-17}~N$
