Find energy of a lightning strike

  • Thread starter Thread starter sfgoat
  • Start date Start date
  • Tags Tags
    Energy Lightning
AI Thread Summary
The discussion focuses on calculating the energy released during a lightning strike, modeled as a parallel-plate capacitor between the Earth and a cloud layer 800 meters above. The capacitance is calculated to be approximately 1.11 x 10^-8 F using the given area and distance. The energy density formula yields an energy release of about 3.19 x 10^10 J, confirming the calculations. An alternative method using the formula U = 1/2 CV^2 is suggested for a more direct approach. The calculations are validated as correct, with both methods being acceptable for determining the energy release.
sfgoat
Messages
10
Reaction score
0
Consider the Earth and the cloud layer 800 meters above the Earth to be the place of a parallel-plate capacitor. The cloud layer has an area of one kilometer squared. Assume this capacitor discharge that is lightning occurs, when the electric field strength between the plates reaches 3.0X10^6 V/m. What is the energy released if the capacitor discharge completely during a lightning strike?

Attempt:
C=(€*A)/d=((8.85*10^-12)(1*10^6m))/800=1.11*10^-8 F
Energy density=(0.5)(€)(E^2)=(.5)(8.85*10^-12)(3*10^6)^2=39.83=energy/volume therefore energy=39.83*(A*800)=3.19*10^10 J

Is this correctly done? Any help is appreciated.
 
Physics news on Phys.org
Yes, that looks right. It's a bit indirect to go through energy density though -- you could solve it a bit more directly with U = \frac{1}{2}CV^{2}. Either is fine though, of course.
 
Thanks.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top