How much negative charge do I accumulate by touching the earth?

Question

The Earth carries a negative electric charge of roughly 500 thousand Coulombs (according to different sources I've seen). If I touch the Earth I should therefore pick up some of this electric charge (through conduction) and become negative charged. Assuming the earth can modeled as a conducting sphere with radius $\small 6371 [\text{km}]$ and me as a conducting sphere with radius $\small 1 \small[\text m]$, around how much negative charge would I accumulate? The reason I ask is because I'm trying to prove to myself that grounding does indeed render a charged object neutral (i.e. transfers all the object's charge to the Earth). Using the well known equation for two connected conducting spheres with different radii (see Example 3-13 on page 115 in David Cheng's "Field and Wave Electromagnetics, 2nd Ed."), I calculate $\small 0.0785 [\text C]$, which is way too big and must be wrong.

Here is my calculation:

$V_\text{sphere}=k\times \frac{Q_1}{r_1} $ (potential of conducting sphere with radius $r_1$ and and net charge $Q_1$) $V_\text{earth}=k\times \frac{Q_2}{r_2}$ (potential of conducting sphere with radius $r_2$ and and net charge $Q_2$)

where $k$ is a constant. If the sphere touches the earth then their potentials ($V_\text{sphere}$ and $V_\text{earth}$) must be equal,assuming that the charges on the spherical conductors may be considered as uniformly disturbed. Setting $V_\text{Sphere}=V_\text{earth}$, we get:

$\frac{Q_1}{r_1}=\frac{Q_2}{r_2}$

Setting $\quad Q_1+Q_2=Q_\text{total}$,

yields:$\quad Q_1=Q_\text{total}\times \frac{r_1}{(r_1+r_2)}$

Substituting

$Q_\text{total}=500,000[\text C],\quad r_1=1 [\text m],\quad r_2=6371000 [\text m]$

I get:

$Q_1=0.0785[\text C]$.

I feel this number is way too large to be correct. If you take coupling into account (by modeling earth as PEC plate, the charge I calculate only gets larger!). What am I doing wrong here? There seems to be no way you accumulate $\small-0.0785 [\text C]$ of charge by touching the earth.

Answer 1

The electric field of earth is really quite complicated. This paper is old, but presents a nice overview of the issues. That $500\,000\, \mathrm{C}$ number is related a charge separation between the surface and the bottom of the ionosphere, rather than the net charge on the earth. And in any case, it is only the result of a rough model whose underpinnings aren't entirely accurate. Your small sphere is embedded inside this massive system (the Earth's atmosphere). As @John Rennie points out in the comments, because the Earth is so large, you can ignore its curvature, and a better model for this system is a small sphere between the two plates of an enormous capacitor with a $\sim 300\,000\, \mathrm{V}$ potential difference.

So, as always, when dealing with potentials, you need to ask yourself: potential relative to what? Well, assuming that the ionosphere is actually at the same potential as a point infinitely far away, the natural interpretation of your question is to assume that your sphere is neutral infinitely far away, is insulated, transported to the surface of the earth, and then touched to the surface of the earth, and ask how much charge is transferred at this last step. My calculations suggest that a $1\,\mathrm{m}$ sphere has a capacitance of about $100\, \mathrm{pF}$. For a voltage difference of $\sim 300\,000\, \mathrm{V}$, that would be a few times $10^{-5}\,\mathrm{C}$.

Incidentally, as my answer on this question suggests, humans are modeled as capacitors with about the same capacitance, so you would get the same amount of charge.

Of course, these calculations are all for isolated objects, which makes them not perfectly valid, but should be reasonable approximations.