How to Find the Potential Energy of a Cloud?

Click For Summary
SUMMARY

The discussion focuses on calculating the potential difference of a cloud using the electric field and distance from Earth. The maximum electric field sustained by dry air is given as 3x10^6 Vm^-1, with a distance of 1000 meters between the Earth and the cloud. The potential difference is calculated using the formula V=Ed, resulting in a value of 8.23x10^11 volts. Participants clarify that the second formula for energy stored in an electric field is unnecessary for this calculation.

PREREQUISITES
  • Understanding of electric fields and potential difference
  • Familiarity with the formula V=Ed
  • Knowledge of the permittivity of free space, ε₀
  • Basic algebra for solving equations
NEXT STEPS
  • Review the concept of electric fields in physics
  • Study the implications of potential difference in electrical systems
  • Learn about the energy stored in electric fields using the formula μ = 1/2(ε₀ × E²)
  • Explore real-world applications of electric fields in meteorology
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism, as well as educators and anyone interested in the electrical properties of clouds.

boneill3
Messages
126
Reaction score
0

Homework Statement



Assuming the maximum electric field sustained by dry air in a cloud is 3x10^6 Vm^-1
And a distanceof 1000 meters between Earth and cloud. The cloud is 4km long and 1 km wide. ausume uniform eletric field. Find the potential difference.

Homework Equations


V=Ed
\mu = \frac{1}{2}(\epsilon_{0}\times E^2)


The Attempt at a Solution



\mu = \frac{1}{2}(\epsilon_{0}\times E^2)
3\times 10^6 = \frac{1}{2}(8.85\times 10^-12\times E^2)
E=\sqrt{\frac{3\times 10^6}{\frac{1}{2}(8.85\times10^-12\timesE^2)}}
E = 8.23\times10^8

using V=Ed
E = 8.23\times10^8 \times 1000
=8.23x10^11 volts


is this right?
 
Last edited:
Physics news on Phys.org
boneill3 said:

Homework Equations


V=Ed
\mu = \frac{1}{2}(\epsilon_{0}\times E^2)

You don't need the second formula (which is for the energy stored in an electric field, I believe. The first equation, along with the given information, should be enough to solve the problem.

boneill3 said:
\mu = \frac{1}{2}(\epsilon_{0}\times E^2)
3\times 10^6 = \frac{1}{2}(8.85\times 10^-12\times E^2)

Why have you plugged in the value of the electric field, E, for mu? The electric field, E, is already given to you. You don't need to solve for it.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
Replies
5
Views
1K
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
7
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K