q = 40.3 microCoulombs
Explanation:The force of attraction between two charges is given by the formula:
[tex]F=\frac{kq^2}{r^2}[/tex]The force of attraction. F = 56.46 N
The separation, r = 50.91 cm
r = 50.91/100
r = 0.5091 m
The electrostatic constant is:
[tex]k=9\times10^9Nm^2C^{-2}[/tex]Solve for the magnitude of charge q
[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ \\ 56.46=\frac{9\times10^9\times q^2}{0.5091^2} \\ \\ 56.46\times0.5091^2=9\times10^9\times q^2 \\ \\ q^2=\frac{56.46\times0.5091^2}{9\times10^9} \\ \\ q^2=1.63\times10^{-9} \\ \\ q=\sqrt{1.63\times10^{-9}} \\ \\ q=4.03\times10^{-5}C \\ \\ q=40.3\mu C \end{gathered}[/tex]