How Is Capacitance Calculated in a Cloud-to-Ground Model?

[Badrakhandama]

1. The base of a cloud has an area of 2*10^7 metres squared and is on average 1km above the ground. treating the base of the cloud, the ground and the intervening air gap as a parallel plate capacitor, calculate the capacitance.[relative permittivity of air = 1.0]





2. I tried to say C = (Eo)(Er)(A)/(D)


And so, C = (8.85*10^-12)(1.0)(2*10^7)/(1000)

And this gives me C = 1.77*10^-7 Farads. However, the answer the book gives is 8.85*10^-9 Farads. Is the book wrong?

 

[madmike159]

I also got 1.77x10^-7 F, so I guess the book must be wrong.

 

[tiny-tim]

Hi Badrakhandama!

(have an epsilon: ε and try using the X² and X₂ icons just above the Reply box )

C = (Eo)(Er)(A)/(D)


And so, C = (8.85*10^-12)(1.0)(2*10^7)/(1000)[/b]

And this gives me C = 1.77*10^-7 Farads. However, the answer the book gives is 8.85*10^-9 Farads. Is the book wrong?

Yes, I can't see anything wrong with your answer either…

the book's answer is exactly 20 times too small, so it looks as if they used A = 10⁶ instead!