How Do You Calculate the Capacitance of a Thundercloud?

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The discussion focuses on calculating the capacitance of a thundercloud with a negative charge of 20 C located 3 km above the Earth's surface. The formula used is C=8.85*(10^(-12))*A/d, where A is the area and d is the separation distance. Participants clarify that the geometry should be treated as a disc rather than a sphere, which affects the calculations. One user initially misapplied the flat plate approximation but later corrected their understanding of the geometry. The correct capacitance value is confirmed to be C=9.3*10^-9 Farads.
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Homework Statement


The charge center of a thundercloud, drifting 3.0km above the Earth's surface, contains 20 C of negative charge.
Assuming the charge center has a radius of 1.0km Calculate the Capacitance of the system.

Homework Equations


C=8.85*(10^(-12))*A/d

Where A is the area, D is the separation distance


The Attempt at a Solution


Basically I just used the formula and plugged and chugged because of the large sizes, we can ignore the curvature and treat it as parallel plate capacitors.
8.85*10^(-12)*4∏(1000)^2/3000

This however gave me the wrong answer, I'm not really sure where to go from here since this path was wrong.

I hope there isn't a limit to how many questions you can ask in a day :P
 
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I don't think the large absolute sizes justify the flat plate approximation. The problem would be essentially the same if the 3km and 1km were replace by 3m and 1m. It's the ratio of these sizes that might justify such an approximation.
What you have here can be thought of either as a sphere and an infinite plate, or as two spheres, non-concentric.
 
haruspex said:
I don't think the large absolute sizes justify the flat plate approximation. The problem would be essentially the same if the 3km and 1km were replace by 3m and 1m. It's the ratio of these sizes that might justify such an approximation.
What you have here can be thought of either as a sphere and an infinite plate, or as two spheres, non-concentric.

I confirmed with my professor about a week ago that it is to be solved using the flat plate approximation. He normally doesn't reply on the weekends though so I figured I'd go on here for help. I can't think of any other path to take regarding this problem.
 
Sorry, I completely misunderstood the geometry. I thought the charge in the cloud was a sphere which you were approximating as flat. I see now it is intended to be a disc. Given that, I can find no flaw in your method. What is the 'correct' answer?
 
haruspex said:
Sorry, I completely misunderstood the geometry. I thought the charge in the cloud was a sphere which you were approximating as flat. I see now it is intended to be a disc. Given that, I can find no flaw in your method. What is the 'correct' answer?

C=9.3*10^-9 Farads.

My answer is 3.7*10^-8, when I punch it in the calculator.
 
So where does the 4 come from in 8.85*10^(-12)*4∏(1000)^2/3000?
 
haruspex said:
So where does the 4 come from in 8.85*10^(-12)*4∏(1000)^2/3000?

Oh no, I was geometrically incorrect, I've been working with so many spheres that I assumed this was a sphere in the math even though I knew it was a disc.

That fixes everything. Thank you!
 
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