Question

In a thunderstorm, 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. For the purposes of this problem, take the cloud to be a sphere of diameter 1.00 kilometer. The point of this problem is to estimate the maximum amount of charge that this cloud can contain, assuming that the charge builds up until the electric field at the surface of the cloud reaches the value at which the surrounding air breaks down. This breakdown means that the air becomes highly ionized, enabling it to conduct the charge from the cloud to the ground or another nearby cloud. The ionized air will then emit light due to the recombination of the electrons and atoms to form excited molecules that radiate light. In addition, the large current will heat up the air, resulting in its rapid expansion. These two phenomena account for the appearance of lightning and the sound of thunder. Take the breakdown electric field of air to be Eb=3.00×106N/C.
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) .
2. Assuming that the cloud is negatively charged, how many excess electrons are on this cloud?

Answer 1

Answer:

1) q = 83.4 C

2)52.13 × 10^(19) electrons

Explanation:

1) To calculate the total charge q on the cloud when the breakdown of the surrounding air begins, we will use the formula;

E = kq/r²

Making q the subject, we have;

q = Er²/k

Where k is a constant = 1/(4πε_o)

We are given;

ε_o = 8.85 × 10^(-12)) C²/N.m²

Thus;

k = 1/(4 × π × 8.85 × 10^(-12)) N.m²/C²

k = 8.99 × 10^(9)

Also,we are given E = 3 × 10^(6) N/C

Diameter = 1km = 1000 m

Radius(r) = diameter/2 = 1000/2 = 500 m

Thus;

q = (3 × 10^(6) × 500²)/(8.99 × 10^(9))

q = 83.4 C

2) To get the number of Excess electrons, we will divide the charge gotten in Part 1 above by the charge of a single electron.

Now, charge of a single electron = 1.6 × 10^(-19) C

Thus, number of Excess elecrons = 83.4/(1.6 × 10^(-19)) = 52.13 × 10^(19) electrons