# Can the Earth's negative charge make negatively charged objects levitate?

Since the earth's surface has a negative charge, could it repel a large highly negatively charged body? An example is an object on stilts carrying a negative charge that is heavier than air, maybe multiple tons in weight.

## 1 Answer

The strength of the electric field near the surface of the Earth due to its negative charge is reported to be around $100 \frac V m$.

This field should easily lift an electron, since the force acting on an electron, $eE=1.6\times10^{-19}C\times 100 \frac V m=1.6\times 10^{-17}N$ is orders of magnitude greater than the weight of an electron $m_e g=9.1\times10^{-31}kG\times 9.8 \frac m {sec^2}=8.9\times10^{-31}N$.

This does not scale favorably, however. As the of an object increases, its mass grows as a cube of its linear dimension, while its charge density as a square.

As a result, for a large object to be lifted by the Earth's electric field, it would have to be charged to an unsustainable voltage level.

As an example, we can consider a conductive sphere with a diameter of $1m$ and assume that it weighs $1kG$ or about $10N$. The charge required to develop $10N$ force in the $100 \frac V m$ electric field, would be $Q=\frac F E=\frac {10N} {100 \frac V m}=0.1C$.

The capacitance of such sphere is $4\pi \epsilon_0R \approx 55pF$, which means that, if it is going to be charged to $0.1C$, its potential will be $V=\frac Q C=\frac {0.1C} {55\times 10^{-12}F}=1.8\times 10^9V=1.8GV$, which is obviously impractical.

Lifting objects "multiple tons in weight" would be even less realistic.

• "As the [length]of an object increases, its mass grows as a cube of its linear dimension, while its charge density as a square", could you please explain why the charge density only grows to the square ? – Manu de Hanoi Sep 19 '18 at 8:35
• @ManudeHanoi Because the charge in a conductor resides on it surface and the surface grows as a square of the linear dimension. – V.F. Sep 19 '18 at 10:35
• May I further ask why does the charge on a conductor reside at its surface instead of being uniformely distributed in its mass ? – Manu de Hanoi Sep 19 '18 at 11:28
• @ManudeHanoi Because excessive charges repel each other and, therefore, move away from each other until they arrive to the surface. – V.F. Sep 19 '18 at 11:36
• but if it was so, then, once at the surface, because the charge density is higher, they would push back charges toward the center right ? So there has to be something else right ? – Manu de Hanoi Sep 19 '18 at 12:51