The electric field of a point charge is given by:
[tex]E=\frac{kq}{r^2}[/tex]For the charge of +2 nC
[tex]E_{+2}=\frac{k(2\times10^{-9}C)}{(8.20^2)}[/tex]For the charge of -4nC:
[tex]E_{-4}=\frac{k(-4\times10^{-9}C)}{(3.23^2)}[/tex]Since the field at the origin must be zero, then:
[tex]\begin{gathered} E_{+2}=E_{-9} \\ so: \\ \frac{k(2\times10^{-9}C)}{(0.082^2)}=\frac{k(-9\times10^{-9})}{y^2} \\ solve_{\text{ }}for_{\text{ }}y: \\ y=\sqrt{\frac{(0.082^2)(-9\times10^{-9})}{(2\times10^{-9})}} \\ y\approx0.1739 \end{gathered}[/tex]And:
[tex]\begin{gathered} E_{-4}=E_{-9} \\ \frac{k(-4\times10^{-9}C)}{(0.0323^2)}=\frac{k(-9\times10^{-9})}{x^2} \\ solve_{\text{ }}for_{\text{ }}x: \\ x=\sqrt{\frac{(0.0323^2)(-9\times10^{-9})}{(-4\times10^{-9})}} \\ x=-0.04845 \end{gathered}[/tex]Therefore, the coordinates of the -9 nC charge are:
(-0.04845,0.1739)
And the distance r is:
[tex]\begin{gathered} r=\sqrt{(-0.04845)^2+(0.1739)^2} \\ r=0.1805cm \end{gathered}[/tex]
