1) 333.6 C
In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:
[tex]E=3.00\cdot 10^6 N/C[/tex]
The electric field strength at the surface of a charged sphere is given by
[tex]E=\frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}[/tex]
where
[tex]\epsilon_0 = 8.85\cdot 10^{-12} C^2/(N^2 m^2)[/tex] is the vacuum permittivity
Q is the charge on the sphere
R is the radius of the sphere
Here we have a cloud of radius
[tex]R = 1.00 km = 1000 m[/tex]
So we can re-arrange the previous equation in order to find the charge on the cloud:
[tex]Q=4\pi \epsilon_0 ER^2=4\pi (8.85\cdot 10^{-12})(3.00\cdot 10^6)(1000)^2=333.6 C[/tex]
2) [tex]2.08\cdot 10^{21}[/tex] excess electrons
The total charge of the cloud must be (in magnitude)
Q = 333.3 C
We know that one electron carries a charge of
[tex]e = 1.6 \cdot 10^{-19}C[/tex]
The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:
[tex]Q=Ne[/tex]
where
N is the number of excess electrons
Solving for N, we find:
[tex]N=\frac{Q}{e}=\frac{333.3 C}{1.6\cdot 10^{-19} C}=2.08\cdot 10^{21}[/tex]
