A negative charge of -6.0*10^-6 C exerts an attractive force of 65N on a second charge .050 m away. What is the magnitude of the second char

Coulomb's law says (1/4*pi*eps)*q*Q/r^2=F  where eps is the permittivity of free space (8.854*10^-12 farad/m) q is the charge on one source, Q is the charge on the other, r is the distance between them and F is the force.  This gives (1/4*pi*8.854*10^-12)(-6*10^-6)*Q/0.05^2=65

$${\mathtt{Q}} = {\frac{{\mathtt{65}}{\mathtt{\,\times\,}}{{\mathtt{0.05}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{8.854}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{12}}\right)}}{{\mathtt{\,-\,}}\left({\mathtt{6}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{-{\mathtt{6}}}\right)}} = {\mathtt{Q}} = -{\mathtt{0.000\: \!003\: \!013\: \!363\: \!313\: \!4}}$$Coulombs or about -3*10-6C