• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Capacitance of a Thundercloud

  • Thread starter mathnoobie
  • Start date
1. The problem statement, all variables and given/known data
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.

2. Relevant equations
C=8.85*(10^(-12))*A/d

Where A is the area, D is the separation distance


3. 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
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,835
4,452
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 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.
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,835
4,452
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?
 
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.
 

haruspex

Science Advisor
Homework Helper
Insights Author
Gold Member
2018 Award
30,835
4,452
So where does the 4 come from in 8.85*10^(-12)*4∏(1000)^2/3000?
 
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!
 

Want to reply to this thread?

"Capacitance of a Thundercloud" You must log in or register to reply here.

Related Threads for: Capacitance of a Thundercloud

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top