Question

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be distributed uniformly throughout the cloud. The charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air "breaks down."
In general, the term "breakdown" refers to the situation when a dielectric (insulator) such as air becomes a conductor. In this case, it means that, because of a very strong electric field, the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air then emits light as the electrons and ionized atoms recombine to form excited molecules that radiate light. The resulting large current heats up the air, causing its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder.

The point of this problem is to estimate the maximum amount of charge that a cloud can contain before breakdown occurs. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 km. Take the breakdown electric field of air to be Eb=3.00×10^6N/C.


Assuming that the cloud is negatively charged, how many excess electrons are on this cloud?

Answer 1

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

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:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$