The radius of the sphere is r=5.15 cm=0.0515 m, and its surface is given by
[tex]A=4 \pi r^2 = 4 \pi (0.0515 m)^2 = 0.033 m^2[/tex]
So the total charge on the surface of the sphere is, using the charge density
[tex]\rho=+1.21 \mu C/m^2 = +1.21 \cdot 10^{-6} C/m^2[/tex]:
[tex]Q= \rho A = (+1.21 \cdot 10^{-6} C/m^2)(0.033 m^2)=4.03 \cdot 10^{-8}C[/tex]
The electrostatic force between the sphere and the point charge is:
[tex]F=k_e \frac{Qq}{r^2} [/tex]
where
ke is the Coulomb's constant
Q is the charge on the sphere
[tex]q=+1.75 \muC = +1.75 \cdot 10^{-6}C[/tex] is the point charge
r is their separation
Re-arranging the equation, we can find the separation between the sphere and the point charge:
[tex]r=\sqrt{ \frac{k_e Q q}{F} }= \sqrt{ \frac{(8.99 \cdot 10^9 Nm^2 C^{-2})(4.03 \cdot 10^{-8} C)(1.75 \cdot 10^{-6}C)}{35.9 \cdot 10^{-3}N} }=0.133 m=13.3 cm [/tex]
