Outside the radius of the sphere, the electric field generated by a charged sphere (with charge Q on its surface) at a distance r from the centre is equivalent to the electric field generated by a single-point charge with total charge Q:
[tex]E= k_e \frac{Q}{r^2} [/tex]
The problem asks to find the electric field at r=8.64 cm, while the radius of the sphere is R=5.00 cm, so r>R and we are exactly in this condition.
Re-arranging the previous formula, we can solve to find the total charge Q on the sphere. Using [tex]r=8.64 cm=8.64 \cdot 10^{-2}m[/tex] and [tex]E=1.53 \cdot 10^5 N/C[/tex], we find
[tex]Q= \frac{Er^2}{k_e} = \frac{(1.53 \cdot 10^5 N/C)(8.64 \cdot 10^{-2}m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} =1.27 \cdot 10^{-7}C[/tex]
This is the total charge on the surface. If we want to find the number of electrons composing this charge, we should divide the total charge by the charge of a single electron, e:
[tex]N= \frac{Q}{e} = \frac{1.27 \cdot 10^{-7}C}{1.6 \cdot 10^{-19}C}=7.9 \cdot 10^{11} [/tex]
and this is the number of electrons.
