Can Activated Carbon Powder Fly Using Atmospheric Electric Field?

  • Context: Graduate 
  • Thread starter Thread starter NERV
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the feasibility of using atmospheric electric fields to levitate activated carbon powder. The atmospheric electric field strength near the ground is approximately 100 N/m, which is sufficient for this application. The author calculates the necessary electric potential (U) to achieve levitation, concluding that a minimum of 1017 V is required, with a practical example of 2034 V providing additional lifting force. The calculations consider the density of activated carbon powder (2.7 g/cm³) and the dimensions of the particles.

PREREQUISITES
  • Understanding of atmospheric electric fields and their strengths
  • Basic knowledge of electrostatics, including charge and electric force
  • Familiarity with the properties of activated carbon, specifically density and conductivity
  • Proficiency in physics calculations involving electric potential and gravitational force
NEXT STEPS
  • Research the principles of electrostatics and electric fields
  • Explore the properties and applications of activated carbon in various fields
  • Learn about the calculations involved in electrostatic levitation
  • Investigate practical experiments involving atmospheric electric fields and particle levitation
USEFUL FOR

Researchers in physics, engineers working on electrostatic applications, and anyone interested in the principles of levitation using electric fields will benefit from this discussion.

NERV
Messages
12
Reaction score
0
As it is known to all,the atmospheric electric field does exists.In fact,near the ground its strength is about 100N/m.That's enough for my design.

Let's get some activated carbon powder.I think we can regard them as smooth,conductive balls with density（ρ）=2.7g/cm3,r=1×10-5m。If we want a can which has been filled with the powder float in the air by balancing the electric force and gravity,we need to charge the powder so that it can hold an amount of electricity.

Here I gave my calculation. I regarded each of these balls as an isolated conductor,which may lead to error.If anyone has some good ideas,please tell me.

C=4πε0r,FE=qE=UCE=4πε0rUE
if a charged ball can fly:
FE≥mg
∴U≥ρgr2/(3ε0E)
Substitute ρ=2.7×103kg/m3,g=10m/s2,r=1×10-5m,ε0=8.85×10-12(C/V.m),E=100N/m,we get

U≥1017V.
This is amazing.If we use U=2034V,we will get one more lifting force(equals the powder's gravity).
 
Physics news on Phys.org

Similar threads

Can Old Handouts Effectively Prepare You for Physics Tests?
  • · Replies 11 ·
Replies
11
Views
4K