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I have a $5 cm×5 cm×? \mu m$ copper plate.

Where '$?$', the thickness of the plate, is really small (about 30um).

Is there anyway using physics to calculate the thickness?

Update:

Sorry I forgot to say that the plate is attached to a non-metallic plate which is bigger than the copper plate from beneath. For instance: its a Printed circuit board.

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  • $\begingroup$ Plenty of ways, but off the shelf mechanical measurement devices are available for that that use mechanical and optical methods. What's wrong with a simple micrometer, for instance? $\endgroup$
    – CuriousOne
    Commented May 19, 2016 at 9:07
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    $\begingroup$ It depends what sort of error you consider acceptable. There are relatively cheap devices for measuring thicknesses on the micrometre length scale, look up Vernier and micrometre calipers. You could also use reticles if you have a microscope. $\endgroup$
    – lemon
    Commented May 19, 2016 at 9:07
  • $\begingroup$ The standard way to do this (say, by PCB manufacturers) is to cross section the part and look at it under a microscope. I would imagine they use a micrometer stage to measure the dimensions accurately, but a reticle or comparison with a reference sample could also be used. $\endgroup$
    – The Photon
    Commented May 20, 2016 at 1:21

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Do you have access to a precision balance? Then you could weigh the plate, and using the known dimensions of the plate and the density of copper, compute the thickness. For $5\,{\rm cm} \times 5\,{\rm cm} \times 30\,\mu{\rm m}$ the weight would be $0.672\,{\rm g}$ for example. The precision of that measurement depends on how accurately you can measure the dimensions, and if the copper is pure or an alloy.

Or you could use a simple micrometer screw gauge.

Edit: As you amended that the copper layer is attached to something, these methods won't work as they are. They would both work if you knew the properties of the thing the copper is attached to. If you have multiple samples and it is possible to destroy one, you could etch away the copper (assuming you can selectively etch the copper, but not the other stuff), and then measure either the weight or the thickness of both the boards with and without the copper and substract them.

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  • $\begingroup$ What if the plate is attached to another substance from one of its biggest faces? $\endgroup$
    – AHB
    Commented May 19, 2016 at 9:34
  • $\begingroup$ @AmirBigdeli you mean that the copper plate is attached to something else on the 5cm x 5cm dimension? That changes everything, you should include details of that in your question, like: Is it possible to separate them? How much do you know about that other substance? $\endgroup$
    – LimaKilo
    Commented May 19, 2016 at 9:59
  • $\begingroup$ It is possible to burn the fiberglass under the copper plate but it will disturb my experiment. I think the best method is to use a micrometer twice. first obtain the fiberglass thickness then over-all thickness then subtract. $\endgroup$
    – AHB
    Commented May 22, 2016 at 5:38
  • $\begingroup$ I think so too, and etching copper from PCB is simple and etching solution readily available. Burning fiberglass would be very messy, also the glass fibers won't burn , only the plastic binder, which might be an additional problem. $\endgroup$
    – LimaKilo
    Commented May 22, 2016 at 8:43
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I calculate that with 30u thickness we have 675mg mass. Use a microbalance and weight the sample.

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If you are reasonably confident about the quality of the plate in question (uniform thickness, a well-defined square shape, etc.) you can measure the resistance of the copper square. This would be a non-destructive method for the arrangement in the OP.

The specific resistivity of copper is (per google search) $1.68 \div 1.72 \times (10^{-8} \Omega\,m)$ (depending if it's annealed or not). Let's take 1.7. Since this is a square piece of copper, only the thickness is relevant.

$$R = \rho / d = 1.7 \times 10^{-2} / d_\mu $$

where $d_\mu$ is the thickness in micrometers and $R$ is in Ohms ($\Omega$). For a $30\,\mathrm{\mu m}$ piece the resistance would be 0.567 milliohms. Small, but quite measurable.

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