Answer
$F = (-4.0\times 10^{-16}~N)~\hat{i}+(1.6\times 10^{-16}~N)~\hat{j}$
Work Step by Step
We can find $E_x$:
$E_x = -\frac{\Delta V}{\Delta x}$
$E_x = -\frac{(-2)(500~V)}{0.4~m}$
$E_x = 2500~V/m$
We can find $F_x$:
$F_x = q~E_x$
$F_x = (-1.6\times 10^{-19}~C)(2500~V/m)$
$F_x = -4.0\times 10^{-16}~N$
We can find $E_y$:
$E_y = -\frac{\Delta V}{\Delta y}$
$E_y = -\frac{400~V}{0.4~m}$
$E_y = -1000~V/m$
We can find $F_y$:
$F_y = q~E_y$
$F_y = (-1.6\times 10^{-19}~C)(-1000~V/m)$
$F_y = 1.6\times 10^{-16}~N$
We can express the force in unit-vector notation:
$F = (-4.0\times 10^{-16}~N)~\hat{i}+(1.6\times 10^{-16}~N)~\hat{j}$
