Solve F=(9*10^{circ}frac{N*m^{2}}{c^2})(2.5*10^-5C)(-3.8*10^-5C)/(0.2m)^2

$$F = ( 9 \times 10 ^ { \circ } \frac { N \cdot m ^ { 2 } } { c ^ { 2 } } ) \frac { ( 2.5 \times 10 ^ { - 5 } C ) ( - 3.8 \times 10 ^ { - 5 } C ) } { ( 0.2 m ) ^ { 2 } }$$

Solve for C

$$\left\{\begin{matrix}C=\frac{4000\sqrt{-\frac{95F}{N}}c}{57}\text{; }C=-\frac{4000\sqrt{-\frac{95F}{N}}c}{57}\text{, }&\left(F\leq 0\text{ or }N<0\right)\text{ and }\left(N>0\text{ or }F\geq 0\right)\text{ and }N\neq 0\text{ and }c\neq 0\text{ and }m\neq 0\\C\in \mathrm{R}\text{, }&F=0\text{ and }N=0\text{ and }c\neq 0\text{ and }m\neq 0\end{matrix}\right.$$

Solve for F

$$F=-\frac{171\times \left(\frac{C}{c}\right)^{2}N}{80000000}$$

$$c\neq 0\text{ and }m\neq 0$$