Suppose I connect the + (or -) pole of a battery to the earth with a conducting wire, will there be a current flow?

As I understand, the Earth is considered to be "neutral" in the sense that it contains inside it many many positive charged particles and negatives particles which cancel each other.

I understand also that if there is a potential difference, then there is a current, but I can't make up my mind because I think that the 2 ends of a circuit should be both "charged" so that the current can flow (like the +/- pole of the battery), but in this case, the positive pole is +, and the earth is neutral.

Can you please help me to clarify this?

  • $\begingroup$ Describe the path that the current would take in the ground. $\endgroup$
    – Bob D
    Mar 16 at 18:58
  • $\begingroup$ @BobD: The current flows from the pole of the battery, through the conductor into the earth ?, as there is a potential difference between the pole and the earth $\endgroup$ Mar 16 at 19:00

2 Answers 2


It doesn't matter whether or not the thing you are connecting is a battery. It doesn't matter whether or not the other thing is the Earth. If you have two conductive objects with no conductive path between them, and if there is a difference of electrical potential between them, then when you create a connection between them (e.g., by attaching a wire) there will be a brief flow of electric current until the potential is equalized.

See also, https://en.wikipedia.org/wiki/Capacitance

A battery is special though because it has two terminals, and it maintains a constant potential between them. Suppose you have a 1.5V battery whose '-' terminal is connected to Earth, and whose '+' terminal is not connected to anything. The '+' terminal must be 1.5V above Earth potential. Now disconnect the wire from the '-' terminal, and connect it to the '+' terminal. A tiny pulse of current must flow to bring the '+' terminal down to Earth potential. At the same time, because of the "constant voltage" property of the battery, that must drive the '-' terminal to be 1.5V below Earth potential.

Now, disconnect the wire once more, and reconnect it to the '-' terminal. Again, a tiny current must flow, in the opposite direction this time, to bring the '-' terminal back up to Earth potential; and that drives the '+' terminal to be 1.5V above Earth potential, bringing the system back to the state where we started.

Repeat as many times as you like, and a tiny pulse of current flows each time.

Something similar happens here, except the Earth is not part of the picture: https://en.wikipedia.org/wiki/Oxford_Electric_Bell

  • $\begingroup$ Hi, so I see that in this case we don't need the closed circuit to have the current, what do you think about it please ? $\endgroup$ Mar 16 at 20:57
  • $\begingroup$ @InTheSearchForKnowledge a circuit is closed when two different potentials are connected through a conductor. Here +(or-) of battery, which is at some potential is connected to earth which is at another potential through a conductor and the circuit is complete. $\endgroup$ Mar 16 at 21:22
  • $\begingroup$ @InTheSearchForKnowledge, You need a closed loop to have continuous, direct (i.e., one-way) current. When you use a single wire to suddenly connect objects that have a potential difference between them, current only will flow until the potential is equalized. If the wire has significant inductance, the current might bounce back and forth a few times, but even that will quickly die out—I would guess, within microseconds in typical cases (for some definition of "typical.") After the potential is equalized, no more current will flow through the connection. $\endgroup$ Mar 16 at 21:43

There will be possibly a very small current, when you connect so the pole of the batterie gets to the same potential as earth. But this has nothing to do with the earth being neutral .

  • $\begingroup$ Hello, can you explain me what do you mean by "the pole gets to the same potential as earth" ? Otherwise, I realize that in my case, the circuit is not closed which means that the wire is not connected to the other end of the battery, is the circuit still valid for current to flow ? If it is, can you explain me why don't we need closed path in this case ? Thanks!! $\endgroup$ Mar 16 at 18:38
  • $\begingroup$ You are right, there is no closed circuit, so you you have no real current , or only if you connect the other pole to earth. What I wanted to say The pole of the batterie and earth may be in a different electric state, called potential, than in the first moment some charges may flow to make the states equal. $\endgroup$
    – trula
    Mar 16 at 18:46
  • $\begingroup$ Hi, there is one thing that confuses me. It is how can we know that the batterie and the earth are in different electric state ? I understand we can simply say that because the voltage of earth is 0. But really, as for me, the earth is a reference point, and if we take the reference point to be another arbitrary point A in the space, the voltage of the earth will change, the voltage of the + pole changes as well, and maybe with respect to the new reference point, they are equal. Do you think can this situation happen ? $\endgroup$ Mar 16 at 18:53
  • $\begingroup$ You use the word voltage wrong. Voltage is all ways between two points. Earth has no "voltage" it ha a potential which we call arbitrarely often 0. We do not know if a pole and earth have the same ore different potential, maybe you brother touched one the pole of the batterie 1h ago to the water pipe, or just by accident they are equal. $\endgroup$
    – trula
    Mar 16 at 20:56
  • $\begingroup$ Hello, in fact, what i wonder is that how can we know that there is a potiential difference between the Earth and the pole of the battery ? At the same time, the potential of a point is defined with respect to a reference point, now I change this reference point (no longer earth), so for example maybe i can find the reference point A so that: potential_battery = potential_earth (both with respect to this point A), then in this case, there is no more potential difference. I'm confused. I don't know if potential difference depends on the reference point or not. Could you please clarify it ? $\endgroup$ Mar 16 at 21:04

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