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I have failed to built a parallel plate capacitor using a pair of Aluminium foils as the parallel plate and air as the dielectric medium. The capacitor is not working at all. My applied potential is around 4V (using power supply) with internal resistance of the power supply is neglected.

Note: The area of the plates is kept at a dimension of 8 cm X 8 cm. The thickness of the plate is around 0.2 mm. The separation between the two plates is 1 cm.

Any advice on how to make sure that my capacitor is working?

Crazy
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  • How are you determining whether the capacitor is working ot not? Have you calculated what the capacitance should be and how much charge you would expect to hold on it at a voltage of 4V? – John Rennie Sep 01 '16 at 12:30
  • Connects the charged capacitor to a voltmeter. (The voltmeter is not showing any deflection which implies no charge is stored). The capacitance should be as in the formulae of C=(kA)/D. To calculate charge. Q= CV – Crazy Sep 01 '16 at 12:31
  • The voltmeter has a finite resistance, so the charge will flow off your capacitor through the voltmeter and be lost. Have you calculated how quickly this will happen? My guess is that you will find it happens so quickly that the voltmeter has no time to react. – John Rennie Sep 01 '16 at 12:35
  • Well. I have tried to increased the potential difference across the capacitor to about 8V. But it still does not show any deflection. Which means that no charge is stored. Are there a better way to make sure that charges are stored in my capacitor. – Crazy Sep 01 '16 at 12:38
  • You'll need to build a Wien bridge or something similar. This is simpler than it sounds, though you'll need a frequency generator. – John Rennie Sep 01 '16 at 12:44
  • If I were to fold the aluminium foils into multiple layers in order to increase the thickness of the parallel plates.(By keeping the area of it to be 8cm square) Would the working formulae be the same? C=(kA)/D – Crazy Sep 01 '16 at 12:48
  • @Crazy - I'm not sure what you mean by $k$ in your expression, but if you mean the conventional $k=\frac{1}{4\pi \epsilon_0}$, then your equation is wrong... – Floris Sep 01 '16 at 12:55
  • it is epsilon x epsilon zero. Not Coulomb's electrostatic force constant. – Crazy Sep 01 '16 at 12:57
  • Although it's on-topic here in my opinion, you might get better answers at electronics.stackexchange.com/ – Martin Sep 01 '16 at 15:37

1 Answers1

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The experiment you describe (in the question and the comments) will almost certainly give you no measurable result; the capacitance of the setup is much too small. Let's do the math:

1 cm spacing, 8 cm x 8 cm area, air gap.

$$C = \frac{\epsilon_0 A}{d} = \frac{8.8 \cdot 10^{-12} \cdot 0.08 \cdot 0.08}{0.01} = 5.6\cdot 10^{-12} F$$

5.6 pF is not a very large capacitance. When you put 4 V across that, you have 20 pC of charge. If your voltmeter has an impedance of 1 MOhm, the RC time constant will be 5.6 µs.

There is no chance that you can observe such a short blip with the naked eye.

You might be able to observe something with a fast oscilloscope, assuming that you do the experiment in vacuum (or at least VERY dry air): the moment you disconnect the voltage source from your capacitor, charge will leak away. Since there is very little charge, a tiny leakage current will be sufficient to remove the charge very quickly.

I addressed the question of charge leakage in this earlier answer - you might find it useful.

You could measure the capacitance of your setup using a (high) frequency generator, and determining that the voltage observed across the capacitor "rolls off" as you increase the frequency. Again, for this you would need a high quality oscilloscope and a variable frequency oscillator. You can look at the phase of the voltage across the capacitor, and you will see that as you reach frequencies corresponding to $\frac{1}{RC}$, the observed phase of the voltage across the capacitor begins to lag.

Your plate has capacitance. Your experiment will not show it.

Community
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Floris
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  • Hmm, I was attempting to get the OP to do this calculation themself ... – John Rennie Sep 01 '16 at 12:50
  • Thanks. I would like to know if the formulae will still remain the same if my foil is folded into multiple layer? The dimension of the folded plates are still 8 X 8. – Crazy Sep 01 '16 at 12:51
  • I am not sure how you plan to fold your foil - but if you increase the area and decrease the distance, you will increase the capacitance. This is why most capacitors are a layered design (which has the added advantage that each side of each plate contributes to the capacitance). – Floris Sep 01 '16 at 12:54
  • The thickness of the capacitor is increased while the area of exposure is remained the same. – Crazy Sep 01 '16 at 12:59