I know potential difference or voltage is meaningful only if we have reference. So I took my reference as the ground and tried to measure the voltage of the 1.5 V battery.

First, I measured $$V(\text{minus side of the battery}) - V(\text{ground})$$ I measured 0 V, or near that.

Then, I measured $$V(\text{plus side of the battery}) - V(\text{ground})$$ I measured the same as the first.

Finally, I derive 0 V as the voltage of the battery. What is the problem here? Shouldn't I get 1.5 V as the result?

edit: my reference is dirt, and yes I actually stuck the reference probe (Black) into the dirt

  • 1
    $\begingroup$ Where is ground (physically) in your measurement? $\endgroup$
    – Bob D
    Jul 24, 2021 at 13:08
  • $\begingroup$ If volt meters could work without completing a circuit, the $0V$ and $0V$ measurements would've shown something like $-\infty V$ and $(1.5 - \infty) V$. $\endgroup$ Jul 24, 2021 at 13:10
  • $\begingroup$ So, you say, if you charge up a metal sphere via friction with a cat fur, You can't measure its potential with reference to ground? $\endgroup$
    – ozgun can
    Jul 24, 2021 at 13:22

4 Answers 4


The issue is that your voltmeter is not perfect. The way a voltmeter normally works is that it consists of a very high resistance (megohm range) between the leads. Then, instead of measuring the voltage difference directly, it measures the current through that path and and solves for the voltage by Ohm's law $V=IR.$

But it only takes a very small amount of charge to change the voltage of an isolated object. A battery maintains the voltage difference between its terminals, but the "absolute" voltage of the terminals (relative to some isolated point) can be changed by adding/removing some net charge to/from the entire battery. (Note that a battery in normal usage in a circuit supports a current, but the net charge of the battery doesn't change with time.)

So what happens is that as soon as you connect two isolated conductors (here the battery and Earth) with your voltmeter, the voltmeter conducts a small current between them as part of its normal operation, but because the conductors are isolated and one of them is small, the tiny charge that flows is enough to quickly change the voltage being measured. In particular, the voltage quickly goes to zero.

Measuring voltages between isolated conductors is basically the realm of static electricity. There exist electrostatic voltmeters that can measure voltages without requiring currents that will alter the measurements. However, static electricity usually involves hundreds/thousands of volts, so such a voltmeter may not register the puny voltage of your battery. I suspect finding components good enough to perform your experiment as stated will be hard.

(Addendum based on Dale's answer: if the connection between your voltmeter and ground through earth is also bad, then effectively your "ground" is another "small conductor" consisting of the bit of soil just around your lead. It now takes even less charge to destroy your measurement!)

  • $\begingroup$ so do you say, In the case of ideal conditions, it would give 1,5V. In the end the problem was with the conditions rather than the theory? $\endgroup$
    – ozgun can
    Jul 24, 2021 at 14:53
  • $\begingroup$ @ozguncan Yes, that sums it up. $\endgroup$
    – HTNW
    Jul 24, 2021 at 14:55
  • $\begingroup$ have 1 more question about a thing partially related to this. When you wire 2 batteries in series ,say 1,5V each, How come you derive 3V potential difference totally. My objection is, when we connect plus side of one battery to the minus side of the other, shouldn't the potential difference between those ends decrease? In the end how come this whole process makes the total voltage 3V?? $\endgroup$
    – ozgun can
    Jul 24, 2021 at 15:09
  • $\begingroup$ @ozguncan If you diagram the four terminals involved as -1 +1 -2 +2 with +1 and -2 connected, then yes, the voltage between +1 and -2 goes to zero quickly on connection. But we don't care about that. We care about the voltage between -1 and +2. The difference between -1 and +1 is 1.5V, and the same between -2 and +2. Now that the +1--2 voltage is zero, you can see the voltage from -1 to +2 is 3V. Before the connection, the -1-+2 voltage could be anything—maybe hundreds of volts, maybe zero. But 3V is what "survives" the surge of equalizing current. $\endgroup$
    – HTNW
    Jul 24, 2021 at 15:14
  • $\begingroup$ guess, the way batteries and power supplies works are different than objects that have static charges. $\endgroup$
    – ozgun can
    Jul 24, 2021 at 16:42

I know Potential difference or voltage is meaningful only if we have reference.

It is potential that is meaningful only if we have a reference. Potential difference does not require a reference. The battery voltage of 1.5 v is the potential difference between the battery terminals.

So if by "ground" you mean that you are physically connecting one of the leads of your voltmeter to the earth or some metal part that is grounded, and the other one of the battery terminals, you will not measure 1.5 volts because the other terminal is open (isolated). To measure the battery voltage, you need to connect your voltmeter leads to the terminals of the battery.

Hope this helps.

  • $\begingroup$ what I dont understand is why the isolation of the non-measured terminal has anything to do with measurement of the other. Electric charge is a quantity that can exist as monopoles unlike in magnetizm. So you can actually charge a conducting sphere via friction and then if you wire it up to the ground, a Current occurs which indicates there is a potential difference. So why is that not the case with the battery terminals? $\endgroup$
    – ozgun can
    Jul 24, 2021 at 13:27
  • $\begingroup$ The battery is isolated from ground unless you connect one of the battery terminals to ground. Any static charge may cause some deflection in the meter, but the meter will not read the battery voltage unless it is connected across the battery. You already proved that to yourself. $\endgroup$
    – Bob D
    Jul 25, 2021 at 1:30

There are a couple of issues with your measurement.

First, simply sticking a little voltmeter probe into dirt is insufficient to “earth” or “ground” your circuit. Typically a grounding rod is around 2 m long and needs to be driven into the ground securely, and if the soil is dry or rocky then you may need to use multiple rods. It is rather exhausting, but you need a fairly large contact area to get a good electrical connection with the earth.

Second, you need to measure the voltages in the same circuit. You measured 0 V in one circuit and then 0 V in a different circuit. The difference is not meaningful. The proper way to do this measurement is to use two voltmeters and measure both simultaneously.

Third, recognize that the voltmeter has a finite impedance. Typically it is very large in comparison to standard circuits, so connecting it only changes the circuit a little. But when the rest of your circuit is an open circuit then the voltmeter itself becomes the low-impedance path. Then the voltmeter substantially alters the circuit.

  • $\begingroup$ What does your second point mean? Is it in reference to the third point (that the voltmeter placement changes the circuit)? (I ask because in that case it is confusing that it comes before.) Or are you implying that even with an ideal voltmeter the OP's experiment would fail? Physically speaking, as long as no charges are flowing, there are two constant voltage differences from the battery terminals to Earth that themselves differ by the battery voltage. Ideal voltmeters can measure them simultaneously or in sequence. There is no reason for that not to work except for non-ideal components. $\endgroup$
    – HTNW
    Jul 24, 2021 at 14:19
  • $\begingroup$ @HTNW I mean that the voltage difference between two points in two different circuits is a meaningless quantity. If I measure the voltage between ground and a point in a radio circuit today and between ground and a point in a computer circuit tomorrow, their difference is meaningless. I can certainly mathematically perform the subtraction, but it doesn’t have any physical meaning. $\endgroup$
    – Dale
    Jul 24, 2021 at 14:32
  • $\begingroup$ @Dale It does have a physical meaning. One of the result of that is, if you were to wire up your circuit of the radio and the computer, A current will occur untill the potential difference between them become zero $\endgroup$
    – ozgun can
    Jul 24, 2021 at 14:38
  • $\begingroup$ @Dale Then I don't see how that's relevant. The voltage you measure between ground and the negative terminal of the battery and the voltage you measure between ground and the positive terminal of the battery are both "meaningless" in the sense that they can in principle "be whatever", but subtract them and the arbitrary part (potential difference between different circuits) cancels out and you should get the "meaningful" voltage between battery terminals. In your case the subtraction gives the voltage between the two points in the two circuits, which is again arbitrary/"meaningless". $\endgroup$
    – HTNW
    Jul 24, 2021 at 14:40
  • $\begingroup$ @Dale it is not the measurement belongs to the two different circuits. it is the measurement of the points which are assumed to be not changed in terms of Amount of charge during and after the measurement. If you claim that The measurement alters the amount of charges of the points significantly , then the whole thing would be meaningless. But in my case, it is not. $\endgroup$
    – ozgun can
    Jul 24, 2021 at 14:46

You start off with the battery and the ground.

The potential difference across the battery terminals is fixed at $1.5\,\rm V$ across its terminals.

Without taking a reading ask yourself, what is the potential difference between each of the terminals and the ground?

You should soon come to the conclusion that with the information given it is indeterminate and for example, the negative terminal could be at a potential of $-1\,\rm V$ relative to the ground and the positive terminal could be at a potential of $+0.5\,\rm V$ relative to the ground or $\pm 0.75 \,\rm V$ or any other combination of potentials as long as the potential difference between the two terminals id $1.5\,\rm V$.
All of this assumes that the ground is an equipotential surface.

So what happens when you connect your voltmeter from the negative terminal to the ground?
If there is a potential difference between the negative terminal and the ground and given that the voltmeter is a conductor, charges will flow until the potential difference between the negative terminal and the ground is zero and then no more current flows and the voltmeter reading becomes zero.
You can then surmise that the potential of the positive terminal relative to the the ground is $+1.5\,\rm V$.

Oh, that means that if I put another voltmeter between the positive and the ground I should get a reading of $+1.5\,\rm V$ on the second voltmeter?

No, that is not the case because you now have a series circuit consisting of two voltmeter, the battery and the ground and the voltmeters are measuring the voltage across their terminals.
If the ground, battery and connecting wires had no resistance and the voltmeters were of the same type then each voltmeter would show a reading of $0.75\,\rm V$.

  • $\begingroup$ I agree until the part that, grounded terminal,say negative one, become equal potential with the ground, but after that, how can the other terminal, in this case positive one, become +1,5V. $\endgroup$
    – ozgun can
    Jul 24, 2021 at 16:40
  • $\begingroup$ If the potential of the negative terminal is zero (= ground) then what is the potential of the positive terminal (relative to ground). $\endgroup$
    – Farcher
    Jul 24, 2021 at 16:58

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