As the problem tells us, the field at point P is 0, thus, the field exerted by charges q1 and q2 have the same magnitude, and exactly the opposite direction, as it can be seen on the following drawing:
Thus, we know that charge q2 will have to be a source of field (as opposed to a sink), and thus, a positive charge. Now all we have to do is find out what charge could produce a field with the same magnitude of the one from q1. As the electric field can be written as:
[tex]E=\frac{kq}{d^2}[/tex]
We'll have:
[tex]\frac{kq_1}{d_1^2}=\frac{kq_2}{d_2^2}\Rightarrow\frac{6.39*10^{-9}}{0.424^2}=\frac{q_2}{0.636^2}\Rightarrow q_2=\frac{6.39*10^{-9}*0.636^2}{0.424^2}[/tex][tex]q_2=1.43775*10^{-8}C[/tex]
Thus, our answer is q2=1.43775*10^(-8)C
Note: Your lesson requests the answer to be inserted as: +1.44
