This is a conceptual question I continue to struggle with as I have started studying electronics.

Say we have a battery, and one of its terminals is connected directly to a nice patch of highly conductive earth. Further assume that the potential of the battery is greater than that of Earth. Now I realize that this is not a closed circuit, but why wouldn't charge flow from the battery to the Earth? Doesn't there exist an electric potential which should cause the quantity of electrons in the battery to be drained or at least decreased to the point at which the electric potential between the Earth and the battery is the same? Isn't this the same principal that is behind electrostatic discharge (albeit this scenario is not implying such a large difference in potential?)

I have read every other answer here on electronics.stackexchange that references ground, and I am still not satisfied.


5 Answers 5


If I understand your question correctly, a battery isn't required to demonstrate this problem. Say you have any object, at some potential. Then, you connect it to some other potential. Does some current flow? Let's say it's a cube of metal, and it's at Earth's potential, plus one volt. Then, it's suddenly connected to Earth:


simulate this circuit – Schematic created using CircuitLab

Short answer: no current flows. There is no circuit for current to flow.

But this is an approximation, made to simplify analysis. We are neglecting an important fact: everything has some capacitance to everything else. The metal cube is one plate of the capacitor, and Earth is the other. So the circuit is actually this:


simulate this circuit

In this case, when V1 suddenly becomes 0V, some current will flow. The total charge that will flow will depend on the capacitance \$C\$, which is very tiny. Maybe \$1fF\$, if even that. We know that capacitance times voltage is charge:

$$ CV = Q $$

So the total charge that will flow if V1 goes from \$1V\$ to \$0V\$, and \$C\$ is \$1fF\$ is:

$$ 1fF \cdot -1V = -1fC $$

This is a very tiny charge, well beyond insignificant to any practical circuit.

The current that will flow is a function of how fast \$V_1\$ changes, and the capacitance \$C\$, according to:

$$ I = C\frac{\mathrm{d}v}{\mathrm{d}t} $$

So how does this relate to ESD?

ESD is what you get when the potential difference between two things is great enough to breakdown the insulation between those things. Usually that insulation is air. How much voltage this takes depends on many factors, and I'm hardly an expert, but we are talking about differences measured in kilovolts.

These high voltages are attainable precisely due to the very small capacitance between you and everything else. Recall again that \$CV=Q\$. We can rearrange that as:

$$ V = \frac{Q}{C} $$

If \$C\$ is very small, then a very small charge \$Q\$ can lead to a very high voltage. When you shuffle across the rug you might only transfer a (metaphoric) handful of electrons, but that's enough to change your voltage relative to your environment quite a lot.

Once you are talking about kilovolts, and not the \$1V\$ in the example as before, that insignificant current is not so insignificant any more. Still small, mind you, but it's applied in such a brief instant it can damage sensitive devices.

Perhaps the most commonly damaged device in modern times is the gate insulation oxide in MOSFETs, which is so thin it might have a breakdown voltage of maybe \$10V\$. If you had enough charge on you to raise your voltage enough to zap the relatively strong air around you, then the few atoms of silicon dioxide can hold that charge back about as well as wet tissue paper:

microscopic ESD damage

  • \$\begingroup\$ So the way that a battery maintains its voltage is by holding constant the static electric fields within its structure such that the line integral of a unit charge traversing from end to end is always the same? But what about an arc discharge? There is no obvious closed circuit in that case, at least not one that I can see. \$\endgroup\$
    – MER
    Commented Jul 12, 2013 at 23:25
  • \$\begingroup\$ @Gigglelot yes. Basically it has to do with the redox potentials of the chemicals in the battery or something, which is constant for a given combination of chemicals, or something. I'm not a chemist, so I don't know exactly how it works. But also, I think I misunderstood your question, so I rewrote it. More insightful now? \$\endgroup\$
    – Phil Frost
    Commented Jul 12, 2013 at 23:30
  • \$\begingroup\$ Ah-ha! That recent edit you made is exactly what I was looking for! I figured it might have been negligible, but none of the resources I had been reading even acknowledged it. \$\endgroup\$
    – MER
    Commented Jul 12, 2013 at 23:32
  • 1
    \$\begingroup\$ @Gigglelot Indeed, you probably wouldn't find this in an electronics textbook. You will find it in physics sources though. There are sources about making useful things, and sources that describe exactly how things actually happen, but unfortunately, never both at the same time :) \$\endgroup\$
    – Phil Frost
    Commented Jul 12, 2013 at 23:40
  • \$\begingroup\$ One last question. I have read that ESD from a person can easily destroy more sensitive components. Does the destruction arise from the "insignificant" charge that flows as described above in your scenario, or is it due to your finger being so large and likely extending across both ends of the device thus completing the circuit? \$\endgroup\$
    – MER
    Commented Jul 13, 2013 at 0:11

Further assume that the potential of the battery is greater than that of Earth.

Batteries are not electrically charged.

If one terminal of a battery is connected to an ideal ground (a perfect sink for electric charge) and charge were to flow out of (or into) that terminal to (from) the ground, the battery would become electrically charged.

But this would increase the potential energy of the system rather than decrease it.*

Another way to see this is that if, say, electrons left the battery, the battery would become positively charged which would attract electrons from the ground back to the battery.

*There may be some minute redistribution of charge depending on the geometry that lowers the energy of the system.

  • \$\begingroup\$ Your hypothetical scenario of charge leaving just to flow back is very elucidating. However, I am still having a hard time reconciling the fact that the battery has no net charge, yet at the same time can supply a voltage. If there is an electric potential difference between the ends of the battery, doesn't this imply a charge gradient from end to end? A gradient implies that a charge at one end will be more positive/negative than a charge at the other end.So I suppose I see that the net charge is zero, but a local derivative of the field would exhibit a net charge. \$\endgroup\$
    – MER
    Commented Jul 12, 2013 at 23:56
  • \$\begingroup\$ @Gigglelot, a battery, via chemical reaction, separates electric charge by removing electrons from one terminal and adding electrons to the other. When an external circuit (closed path) is connected to the battery, the chemical reaction can proceed because the separated charge flows from one terminal to the other via that circuit. If there is no external circuit, any charge that flows from or to the battery leaves the battery with a net electrical charge. \$\endgroup\$ Commented Jul 13, 2013 at 0:00

Isn't this the same principal that is behind electrostatic discharge (albeit this scenario is not implying such a large difference in potential?)

Electrostatic discharge can only happen if there is sufficient potential to break across (or through) the barrier imposed by air (or vacuum or some other gas).

Here is an explanation of Paschen's law. This relates to the terminal voltage required versus the "gap" between terminals for an electric arc to cause current flow at various gas pressures: -

enter image description here

Note that if your battery is below 100V, even at the optimum pressure for the most optimum gas (Argon) you'd struggle to get current to flow. However, if your battery terminals were shaped optimumly there's a better chance of getting current to start flowing. I'm not going to go down that route in this answer unless requested.

It doesn't matter if your battery is earthed/grounded at one terminal or not - it's the potential difference (aka voltage) across the battery that determines whether it discharges through the air/gas/vacuum.


This question is the same question as mine, when I begin to studying electronics.

  1. Further assume that the potential of the battery is greater than that of Earth, why there is no current flow when we connect the single terminal in the battery to the ground?

Simple answer, because the battery is a galvanic cell which needs a chemical reaction to allow current to flow from its terminals. Without connection from + and - terminals, there is no current flowing there.

The battery is not like lightning that has a different potential (sky voltage) between the air and the ground (earth, soil).


The battery has no net charge. Even in a closed circuit the net charge on the battery is zero. However when the battery is connected in a closed circuit, meaning that a conductive path has been established between the (+) and (-) terminals of the battery, charge carries are pumped through the conductive path between the batteries terminals because there is a potential difference between the terminals.

Connecting a single terminal of an ordinary battery to earth's conductive dirt would be the same as connecting a conductive object that has no net charge to ground; nothing happens. Connecting only one terminal to ground is the same as storing the battery in a wooden drawer (except leaving the terminal in the ground will likely corrode it)

If you were to connect both terminal to the conductive dirt in the earth, then you have completed the circuit and charge will flow through the ground between the battery's terminals.

If say you had a battery with 3000 V potential difference between its terminals, and connected one terminal to earth's conductive dirt and maintained a maximum air gap of 1 mm between the other terminal and the ground, the air will likely break down and complete the circuit (battery terminal, earth, ionized air, other battery terminal) allowing current to flow between the terminals of the battery.


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