Answer:
∈=[tex]3.1584x10^{26} \frac{V}{m}[/tex]
Explanation:
Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus
∈[tex]=\frac{P}{E_{o}}[/tex]
The P is the charge density and 'Eo' is the constant of permittivity in free space
to find P
[tex]P=\frac{q}{V}[/tex]
[tex]V=\frac{4}{3}*\pi*r^3[/tex]
[tex]V=\frac{4}{3}\pi*(7x10^{-15})^3[/tex]
[tex]V=2.932x10^{-14} m^3[/tex]
[tex]P=\frac{82C}{2.932x10^{-14}m^3}=2.7965x10^{15} \frac{C}{m^3}[/tex]
So replacing
∈[tex]=\frac{2.7965x10^{15}\frac{C}{m^3}}{8.8542x10^{-12}\frac{C^2}{N*m^2}}[/tex]
∈=[tex]3.1584x10^{26} \frac{V}{m}[/tex]