With a voltmeter , I tested an 18V battery (9+9). It gave 18V. Then I took 2 aluminium square pieces and stuck them on 2 sides of wet flour. After 15 minutes the flour dried(no conduction). This way I made a capacitor. Then I tested the capacitor after connecting it's 2 plates (aluminium foils) to the voltmeter and it showed 0V. Now the problem. I charged the capacitor with the 18V battery for 3 minutes (later I charged it for 8 minutes but the results I will mention remained the same ). After charging , I connected it to Voltmeter and it showed 1V I allowed the capacitor to discharge for 5 minutes but the reading was stable at 1V even after 5 minutes. I took out the foils touched them for a 10 seconds or so and then reconnected to the Voltmeter and the needle was just deflected giving a reading below 0.5V (below least count). Questions Now isn't 3 minutes a lot to charge the capacitor to 18V (the battery was 18V and the wires were simply small ) ? Why did it always charge to 1V only ? It should have been charged to 18 V within seconds. Then why it didn't ? Secondly after 5 minutes of discharge it's voltage was still 1V. How can that be ? It should have been discharged within seconds. Even after touching the plates it was still not at 0. How can that be ? Why did it not charge to 18V and why did it not discharge ?

  • $\begingroup$ It makes no sense, "sticking aluminum to wet floors". Also give a drawing. see this youtube.com/watch?v=uC7HN9oFpZ0 $\endgroup$ – anna v Nov 22 '16 at 17:28
  • $\begingroup$ Your homemade capacitor is faulty. Current runs through it when charging. $\endgroup$ – Gert Nov 22 '16 at 18:08
  • $\begingroup$ @Gert Ok , even if you say that the flour is wet and current runs through it and that's why it doesn't share to 18 V but then what causes it not discharge $\endgroup$ – Shashaank Nov 22 '16 at 18:13
  • $\begingroup$ @annav So according to you , if I make this capacitor with paper as dielectric , then it should be charged to 18V and discharge instantaneously. $\endgroup$ – Shashaank Nov 22 '16 at 18:21
  • $\begingroup$ schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/… $\endgroup$ – anna v Nov 22 '16 at 19:24

You cannot measure the voltage of a typical capacitance with a typical voltmeter due to the voltage change of the source caused by the internal resistance of the voltmeter. Typical voltmeters (multimeters) have an input resistance of about $R_i=10 M \Omega$. Therefore, to measure a voltage $V$, a current $I=V_i/R_i$ has to flow during the measurement time of a couple of seconds without changing the voltage of the source. If you have an ideal capacitor with capacitance $C$ charged to a voltage $V$ and connect it to the voltmeter, the typical discharge time constant is $\tau=R_iC$. With a typical capacitance of such a set-up of several picofarad ($10^{-12}F$), the time constant is on the order of $\tau ∼10^{6}\Omega·10^{-12}F=10^{-6} s$, which is orders of magnitude shorter than the response time of the multimeter, and of you eye. So when you connect your multimeter to such a capacitor, the capacitor is discharged when the voltage meter gives a reading and the voltage reading is zero.

Therefore, in this experiment, you are measuring something not related to capacitance. In essence you have a galvanic cell (even when you think that the flour is "dry") and by "charging" it you induce chemical changes at the aluminum electrodes, so that after applying the battery for some time the electrodes are chemically different and you have a galvanic cell with a measurable voltage.

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  • $\begingroup$ The first sentence is, as written, misleading if not false. An ideal capacitor has infinite parallel resistance but zero series resistance thus, there is no voltage division as implied. $\endgroup$ – Alfred Centauri Nov 22 '16 at 18:07
  • $\begingroup$ Then what is that Voltmeter is measuring $\endgroup$ – Shashaank Nov 22 '16 at 18:14
  • $\begingroup$ @Shashank - You are measuring a galvanic voltage, as I explained in my answer. See en.wikipedia.org/wiki/Galvanic_cell $\endgroup$ – freecharly Nov 22 '16 at 18:39
  • $\begingroup$ @freecharly Ok , I understand bu Would a normal Voltmeter not be able to give the correct potential across a capacitor (good one , working factory made)? $\endgroup$ – Shashaank Nov 22 '16 at 19:18
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    $\begingroup$ @Shashaank- Your findings are a very nice demonstration of the galvanic origin of the observed effects. I wish you further success in your studies and lots of more fun as a physicist! $\endgroup$ – freecharly Nov 23 '16 at 19:20

I don't wish to address the experimental results given but instead, point out a potential flaw in the experiment.

Assuming that you've made a capacitor, i.e., that there isn't some conductive path between the plates, it is likely that the capacitance of your capacitor is extremely small.

For example, assume the area of your aluminum squares is $0.01 \mathrm{m}^2$ and the spacing is nominally $0.001\mathrm{m}$. Then, assuming for simplicity $\epsilon_r = 1$, the capacitance is roughly

$$C = \frac{\epsilon_0 A}{d} = \frac{8.85\times10^{-12} \cdot 0.01}{0.001} \approx 100 \mathrm{pF}$$

If we assume that your voltmeter input resistance is, e.g., $10\mathrm{M\Omega}$, the measurement time constant is

$$\tau = R_{in}C = 10 \times 10^6 \cdot 100 \times 10^{-12} = 1 \mathrm{ms}$$

That is to say, assuming you have built a capacitor in fact and assuming it is charged to some voltage, when you connect your voltmeter across the charged capacitor, the capacitor will discharge through the input resistance in about

$$t_d = 5\cdot\tau = 5 \mathrm{ms}$$

If you want to be able to measure the voltage across the charged capacitor with a voltmeter, you'll need for the capacitance to be large enough such that the time constant is of the order of seconds which requires roughly $10,000$ times larger capacitance.

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  • $\begingroup$ Buy then how come the Voltmeter is measuring 1V and the reading is stable $\endgroup$ – Shashaank Nov 22 '16 at 18:16
  • $\begingroup$ @Shashaank, I quote from my first sentence: "I don't wish to address the experimental results". $\endgroup$ – Alfred Centauri Nov 22 '16 at 18:38

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