A charged particle with a charge of -2.0 × 10^ -6 C is positioned...

The electric field produced by a large nonconducting plate with surface charge density $\sigma$ is given by:

$E = \dfrac{\sigma}{2\epsilon_0}$

The force experienced by a charged particle of charge $q$ in this electric field is:

$F = |q|E = \dfrac{|q|\sigma}{2\epsilon_0}$

Substituting the given values $|q| = 2.0 \times 10^{-6} \text{ C}$, $\sigma = 4.0 \times 10^{-6} \text{ C m}^{-2}$, and $\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$:

$F = \dfrac{2.0 \times 10^{-6} \times 4.0 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

$F = \dfrac{8.0 \times 10^{-12}}{17.7 \times 10^{-12}}$

$F \approx 0.45 \text{ N}$