The magnitude E of the electrical field created by a charge particle with charge q at a distance r away, is:
[tex]E=\frac{kq}{r^2}[/tex]Where k is the Coulomb constant which has a value:
[tex]k=8.99\times10^9N\cdot\frac{m^2}{C^2}[/tex]Substitute q=4.95*10^-4 C and r=7.01*10^-2 m to find the magnitude of the electrical field:
[tex]\begin{gathered} E=\frac{kq}{r^2} \\ =\frac{(8.99\times10^9N\cdot\frac{m^2}{C^2})(4.95\times10^{-4}C)^{}}{(7.01\times10^{-2}m)^2} \\ =\frac{4.45\times10^6\frac{N}{C}m^2}{4.91\times10^{-3}m^2} \\ =9.06\times10^8\cdot\frac{N}{C} \end{gathered}[/tex]Therefore, the answer is option b:
[tex]9.08\times10^6\frac{N}{C}[/tex]