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=3.00106N/C .

question 1:

Estimate the total charge q on the cloud when the breakdown of the surrounding air begins.

Express your answer numerically in coulombs, to three significant figures, using ?0=8.8510?12C2/(N?m2) .

Question 2:

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

Answer 1

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:

$E=3.00\cdot 10^6 N/C$

The electric field strength at the surface of a charged sphere is given by

$E=\frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} C^2/(N^2 m^2)$ 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

$R = 1.00 km = 1000 m$

So we can re-arrange the previous equation in order to find the charge on the cloud:

$Q=4\pi \epsilon_0 ER^2=4\pi (8.85\cdot 10^{-12})(3.00\cdot 10^6)(1000)^2=333.6 C$

2) $2.08\cdot 10^{21}$ 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

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

The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:

$Q=Ne$

where

N is the number of excess electrons

Solving for N, we find:

$N=\frac{Q}{e}=\frac{333.3 C}{1.6\cdot 10^{-19} C}=2.08\cdot 10^{21}$