Answer:
As we know that all conductors are equipotential so we can say that all points lie on the surface of conductors will be at same potential.
So here we can say that
[tex]V = \frac{kQ}{R}[/tex]
[tex]V = \frac{Q}{4\pi \epsilon_0 R}[/tex]
now we will multiply numerator and denominator by radius R
so here we will have
[tex]V = \frac{QR}{4\pi R^2 \epsilon_0}[/tex]
now we know that
[tex]\sigma = \frac{Q}{4\pi R^2}[/tex]
so we have
[tex]V = \frac{\sigma R}{\epsilon_0}[/tex]
since potential is constant so here charge density and radius product will remain constant always
so here on spherical conductor the charge density will remain same at all points
