Answer:
55.6 nC
Explanation:
The electric field at the surface of a charged sphere has the same expression of the electric field produced by a single point charge located at the centre of the sphere and having the same charge of the sphere, so it is given by
[tex]E=k\frac{Q}{r^2}[/tex]
where
[tex]k=9\cdot 10^9 N m^2 C^{-2}[/tex] is the Coulomb's constant
Q is the charge on the sphere
r is the radius of the sphere
In this problem we know
E = 890 N/C is the magnitude of the electric field on the sphere
r = 0.750 m is the radius of the sphere
So by re-arranging the equation we can find the net charge on the sphere:
[tex]Q=\frac{Er^2}{k}=\frac{(890)(0.750)^2}{9\cdot 10^9}=5.56\cdot 10^{-8} C=55.6 nC[/tex]
