Answer:
[tex]2.1\cdot 10^{21}[/tex] electrons
Explanation:
The magnitude of the electric field outside an electrically charged sphere is given by the equation
[tex]E=\frac{kQ}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge stored on the sphere
r is the distance (from the centre of the sphere) at which the field is calculated
In this problem, the cloud is assumed to be a charged sphere, so we have:
[tex]E_b=3.00\cdot 10^6 N/C[/tex] is the maximum electric field strength tolerated by the air before breakdown occurs
[tex]r=1.00 km = 1000 m[/tex] is the radius of the sphere
Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:
[tex]Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C[/tex]
Assuming that the cloud is negatively charged, then
[tex]Q=-333.3 C[/tex]
And since the charge of one electron is
[tex]e=-1.6\cdot 10^{-19}C[/tex]
The number of excess electrons on the cloud is
[tex]N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}[/tex]