(b) [tex]1.54\cdot 10^{-5} C[/tex]
We can actually solve first part (b) the problem. In fact, we know that the electric field strength at the surface of the a sphere is given by
[tex]E=\frac{kQ}{R^2}[/tex]
where
k is the Coulomb's constant
Q is the charge on the surface of the sphere
R is the radius
For this sphere, the radius is half the diameter, so
[tex]R=\frac{43.0 cm}{2}=21.5 cm = 0.215 m[/tex]
We also know that the maximum charge is the value of charge deposited at which the electric field of the sphere becomes equal to the breakdown electric field, so
[tex]E=3.00 \cdot 10^6 V/m[/tex]
Solving the formula for Q, we find the maximum charge:
[tex]Q=\frac{ER^2}{k}=\frac{(3.00\cdot 10^6)(0.215)^2}{9\cdot 10^9}=1.54\cdot 10^{-5} C[/tex]
(a) [tex]6.45\cdot 10^5 V[/tex]
The maximum potential of a charged sphere occurs at the surface of the sphere, and it is given by
[tex]V=\frac{kQ}{R}[/tex]
where we already found at point b)
[tex]Q=1.54\cdot 10^{-5} C[/tex]
and we know that
R = 0.215 m
Solving for V, we find:
[tex]V=\frac{(9\cdot 10^9)(1.54\cdot 10^{-5})}{0.215}=6.45\cdot 10^5 V[/tex]
