Answer
The kinetic energy increased $2.4\times 10^{-19}~J$
Work Step by Step
We can use conservation of energy to find the change in kinetic energy $\Delta K$:
$K_2+U_2 = K_1+U_1$
$K_2-K_1 = U_1-U_2$
$\Delta K = -\Delta V~q$
$\Delta K = -E~d~q$
$\Delta K = -(500.0~N/C)(3.0\times 10^{-3}~m)(-1.6\times 10^{-19}~C)$
$\Delta K = 2.4\times 10^{-19}~J$
The kinetic energy increased $2.4\times 10^{-19}~J$.
