Regarding the Earth and a cloud layer 750 m above the Earth as the "plates" of a capacitor, calculate the capacitance of the Earth–cloud layer system.

(a) Regarding the Earth and a cloud layer 750 m above the Earth as the "plates" of a capacitor, calculate the capacitance of the Earth–cloud layer system. Assume the cloud layer has an area of 1.00 km2 and the air between the cloud and the ground is pure and dry. Assume charge builds up on the cloud and on the ground until a uniform electric field of

4.00 106 N/C throughout the space between them makes the air break down and conduct electricity as a lightning bolt.
nF

(b) What is the maximum charge the cloud can hold?
C
asked Jan 27, 2012 in Physics

In this question, we should use the formula for Capacitance, where

C = k*E*A / d      (*where k = kappa of air, E = epsilon = permitivity of free space, A = area of cloud layer, d = distance )

k = 1

E = 8.85 * 10^-12 C^2 / N*m^2

A  = (1*10^3 m)^2

d = 750 m

Now plug in these values into the formula:

C = 1 * (8.85 * 10^-12 C^2 / N*m^2) * ((1*10^3 m)^2  ) / (750 m)

so C = 1.18*10^-8 F , which is equal to

C = 11.8 nF
answered Jan 28, 2012 by ~Expert~ (3,856 points)
selected Jan 28, 2012 by Roar3215
Part B:

To find the max charge, we use the formula for capacitance

The formula for Capacitance is:

C = Q / V  (where Q = charge, and V = voltage)

Rearrange, the equation so you are solving for Q.

Q = C*V

Now we just need to find V,

V = E*d (where E = electric field and d = distance )

E = 4 * 10^6 N/C

d = 750 m

V = 4*10^6 N/C * 750 m = 3*10^9 V

Now plug back in the equation for the charge, using C = 1.18 *10^-8 F which was calculated in Part A.

and so

Q = (-1.18*10^-8 F)  * (3*10^9 V) = 35.4 C