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.


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.

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  • $\begingroup$ "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 ? $\endgroup$ – Manu de Hanoi Sep 19 '18 at 8:35
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    $\begingroup$ @ManudeHanoi Because the charge in a conductor resides on it surface and the surface grows as a square of the linear dimension. $\endgroup$ – V.F. Sep 19 '18 at 10:35
  • $\begingroup$ May I further ask why does the charge on a conductor reside at its surface instead of being uniformely distributed in its mass ? $\endgroup$ – Manu de Hanoi Sep 19 '18 at 11:28
  • $\begingroup$ @ManudeHanoi Because excessive charges repel each other and, therefore, move away from each other until they arrive to the surface. $\endgroup$ – V.F. Sep 19 '18 at 11:36
  • $\begingroup$ 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 ? $\endgroup$ – Manu de Hanoi Sep 19 '18 at 12:51

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