- #1
NERV
- 12
- 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).
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).