Answer:
Part a)
[tex]Q = 4\pi \epsilon_0 r V[/tex]
Part b)
[tex]\sigma = \frac{\epsilon_0 V}{r}[/tex]
Explanation:
Part a)
As we know that potential of the sphere is given as
[tex]V = \frac{kQ}{r}[/tex]
[tex]V = \frac{Q}{4\pi \epsilon_0 r}[/tex]
now we have
[tex]Q = 4\pi \epsilon_0 r V[/tex]
Part b)
Now we know that for conducting charged sphere whole charge is distributed over the surface of the sphere
so we will have
[tex]\sigma = \frac{Q}{4\pi r^2}[/tex]
now we have
[tex]\sigma = \frac{4\pi \epsilon_0 r V}{4\pi r^2}[/tex]
[tex]\sigma = \frac{\epsilon_0 V}{r}[/tex]