Let's say I have a battery and a capacitor that is neutral. Now I connect the battery positive terminal to only one plate of the capacitor. From what I know, there is a potential difference between the terminal and the plate so there should be an electric field that causes some of the negative charges of the capacitor connected plate to move to the terminal (through a conducting wire) to reduce the potential difference to zero. This makes the connected capacitor plate to be positively charged which in turn causes the not connected plate to be negatively charged because there is an electric field between the plates that causes negative charges to gather in the not connected plate. Now after this happens I expect that there is a difference in potential between the plates of capacitor. But with a multimeter connected to the plates of the capacitor I can observe zero voltage. Why is that? What do I miss?
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1$\begingroup$ What is the negative terminal of the battery connected to? $\endgroup$– FarcherCommented Nov 4, 2022 at 15:04
2 Answers
There is no potential difference because the connection of the positive battery terminal simply causes a redistribution of charge on both plates, but no net charge on either plate. That would require current in a complete circuit involving the other battery terminal.
The free electrons of the connected plate move towards the surface of the plate connected to the positive battery terminal. That, in turn, induces movement of free electrons on the non connected plate towards the surface nearest the connected plate. But the end result is the net charge on the two plates remains zero for a potential difference of zero between the plates.
See FIG 1 below.
Hope this helps.
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$\begingroup$ Oh, I get it. But now I tried something else. I have measured voltage between the positive battery terminal and a neutral wire (not connected to anything), but it still shows zero voltage. I don't know why, because there is a positive net charge in the terminal and zero net charge in the wire, so there should be a potential difference. $\endgroup$– IkarugaCommented Nov 4, 2022 at 14:29
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1$\begingroup$ Why should there be a potential difference? A battery makes a potential difference between its positive and negative terminal. It does not, by itself, create a potential difference involving any other body. $\endgroup$ Commented Nov 4, 2022 at 14:36
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$\begingroup$ Sorry, I can't edit my previous comment. When now I think about this, the connected plate has positive charge and the non connected plate has the net charge of zero, so shouldn't there be a non-zero voltage between the plates anyway? $\endgroup$– IkarugaCommented Nov 4, 2022 at 14:38
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$\begingroup$ @JohnDoty from what I know, when there is a non-zero charge then there is electric potential. I assumed that the positive battery terminal has a positive charge which creates this potential. So when measured potential between positive and neutral charge shouldn't there be a potential difference equal the positive charge value? $\endgroup$– IkarugaCommented Nov 4, 2022 at 14:44
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$\begingroup$ @Ikaruga There is a potential difference between the positive terminal and a point in space (neutral wire or otherwise) but your voltmeter isn't capable of measuring it. That's because all voltmeters need to draw some, albeit small, current from the circuit. If that circuit includes air with its extremely high resistance, the current drawn by the voltmeter may be too small to cause a deflection of the needle (for an analog VM). $\endgroup$– Bob DCommented Nov 4, 2022 at 15:48
You have not mentioned anything about the negative terminal of the battery so I have not connected it to anything.
In the diagrams below the positions of charges and there distribution are illustrative as I do not know exactly how the positive charges will be distributed in diagram $\bf b$.
Diagram $\bf a$ shows the separation of charges on the terminals of the battery due to the electro-chemical process within the battery.
Diagram $\bf b$ shows the connection of the positive terminal of the battery to one plate of the capacitor.
Note that the bottom plate of the capacitor has to stay electrically neutral and not negatively charged as you have stated.
Diagram $\bf c$ shows a voltmeter connected across the plates of the capacitor which leads to the capacitor discharging (inner charges neutralise one another) and the reading on the voltmeter showing zero as in the final state (after more than five time constants $R_{\rm voltmeter}C_{\rm capacitor})$, effectively no current is passing through the voltmeter and so its reading is zero.
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$\begingroup$ @SolomonSlow Thanks, I have incorporated your perceptive comment into my answer. $\endgroup$– FarcherCommented Nov 4, 2022 at 17:41