Answer
$$3.60 \times 10^{-6} \mathrm{N}$$
Work Step by Step
since $$q_{A}=-2.00 \mathrm{nC}$$ and $$q_{C}=+8.00 \mathrm{nC},$$ and bu using Eq. $21-4$ which leads to
$$\left|\vec{F}_{A C}\right|=\frac{\left|q_{A} q_{C}\right|}{4 \pi \varepsilon_{0} d^{2}} $$ $$=\frac{\left|\left(8.99 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(-2.00 \times 10^{-9} \mathrm{C}\right)\left(8.00 \times 10^{-9} \mathrm{C}\right)\right|}{(0.200 \mathrm{m})^{2}}$$ $$=3.60 \times 10^{-6} \mathrm{N}$$
