Answer
The magnitude of the electric force on the electron is $~~1.6\times 10^{-17}~N$
Work Step by Step
We can find the strength of the electric field between $x = 5.0~cm$ and $x = 6.0~cm$:
$E = -\frac{\Delta V}{\Delta x}$
$E = -\frac{5.0~V-6.0~V}{0.060~m-0.050~m}$
$E = 100~V/m$
Since the electric field points from higher potential to lower potential, the electric field is in the +x direction.
Note that the charge on an electron is negative.
We can find the force on the electron:
$F = q~E$
$F = (-1.6\times 10^{-19}~C)(100~V/m)$
$F = -1.6\times 10^{-17}~N$
The magnitude of the electric force on the electron is $~~1.6\times 10^{-17}~N$.
