Find energy of a lightning strike

  • Thread starter Thread starter sfgoat
  • Start date Start date
  • Tags Tags
    Energy Lightning
Click For Summary
SUMMARY

The energy released during a lightning strike can be calculated using the concept of a parallel-plate capacitor, with the cloud layer acting as one plate and the Earth as the other. Given an electric field strength of 3.0 x 10^6 V/m and a capacitor area of 1 km², the capacitance is determined to be 1.11 x 10^-8 F. The energy density formula yields an energy release of approximately 3.19 x 10^10 J. An alternative method using the formula U = 1/2 CV² is also valid for this calculation.

PREREQUISITES
  • Understanding of capacitor theory and formulas
  • Knowledge of electric field strength and its implications
  • Familiarity with energy density calculations
  • Basic grasp of physics principles related to electricity
NEXT STEPS
  • Research the derivation and applications of the capacitor energy formula U = 1/2 CV²
  • Explore the physics of electric fields and their measurement
  • Investigate the implications of energy density in various physical systems
  • Learn about the characteristics and behavior of lightning strikes in meteorology
USEFUL FOR

Physics students, electrical engineers, meteorologists, and anyone interested in understanding the energy dynamics of lightning strikes.

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.
 

Similar threads

Replies
14
Views
4K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K