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Problem Meowtivation and Purrpose

How does one measure weight in space? Certainly not with a scale, because there's no gravity. One must use a special apparatus to deduce it indirectly - through oscillation.

Similarly, you are trying to measure a value of a cat, whereby you cannot directly measure the capacitance. Luckily, there are a few things we know from physics that occur in capacitors that we can use to deduce our feline Faradicity.

Geomeowtry

Let's start by examining the geometry of this problem. We can't exactly state that the cat is a capacitor in the traditional sense, though it can certainly store charge. Practically, you have described a combination floor-paw-cat system, whereby the cat's paws form a dielectric between it and the floor (or bed, or sheets, or whatever). The cat is just one half of the setup, but I digress.

We will thus avoid taking such drastic measures as frying the cat with 10,000 V from head-to-tail (we already know we can model a cat as a resistor). Instead, we will do something fairly harmless: stick the cat on an insulating mat (just for safety) and pull 10,000 V from cat-to-ground.

What happens when a body stores charge?

More charge = more energy. More energy = more mass.

More charge = more ions. More ions = more force somewhere.

Looks like we have two different ways we can make a simple measurement.

Meowthed 1: More charge, more mass

Let's do some napkin-derivation from this brilliant revelation from Einstein.

$$ \begin{split} E &= mc^2 \\ m &= \frac{E}{c^2} \quad\text{a little rearrangement} \\ \partial m &= \frac{\partial E}{c^2} \quad\text{convert into differential form} \end{split} $$

Okay, whatever, where am I going with this? Do you see it? We can now relate a change in mass with a change in energy! That nefarious E term isn't so scary, it's equivalent to the amount of energy stored in the catpacitor.

$$ E_{joules} = C \cdot V \quad \text{(coloumb volts)} \\ 1 C = 1 F \cdot 1 V \\ \therefore E_{joules} = F \cdot V^2 $$

Now we're getting there. Let's combine! $$ \begin{split} \partial m &= \frac{\partial[ E ]}{c^2}, \quad E = F \cdot V^2 \\ \partial m &= \frac{\partial[ F \cdot V^2 ] }{c^2} \\ &= \frac{F}{c^2} \partial [ V^2 ] \\ F &= \frac{ \partial m \cdot c^2 }{\partial [ V^2 ] } \end{split} $$

There you have it, my friend - a formula for the capacitance of a cat that you can measure with a household scale and a voltage source - maybe about a thousand 9V batteries in series. Let's give it a try. Assuming cats are similar to humans, we can estimate the capacitance at around 100 pF. Let's see what to expect at 10,000 V one megavolt.

$$ 100 \text{pF} = \frac{ \partial m \cdot c^2 }{ [ 10^6 \text{V} ]^2 }, \quad \partial m \Rightarrow 1.11 \text{fg} $$

Well, if you must complain about something, it is true that we might miss the change in mass from the breathing of the cat or the normal shedding of fur/skin. Also, we might arc across the insulating mat at one million volts, but hey - you wanted something easy to measure, and what is easier than weighing a cat?

Meowthed 2: More charge, more force

We need two levels of indirection for this one because force can be tricky to measure when it's small (see above). Although we could use another scale with the cat on it, let's rely on something simple - the fact that cats always.land.on.their.feet.

This does require some equipment, namely some big magnets. Take our test platform from meowthed one (the cat, the mat, and the ground plane) and drop them together through the magnets.

$$ \vec F = q(\vec E + \vec v \times \vec B) $$

We can start by eliminating the electric field because we haven't specifically created one. Then, note that the charge we are dealing with comes from the capacitance of the cat.

$$ \begin{split} C &= \frac{q}{V} \quad q = CV \\ \vec F &= CV ( \vec v \times \vec B ) \\ C &= \Big ( \frac{1}{V} \Big ) \Big ( \frac{ \vec F }{ \vec v \times \vec B } \Big ) \end{split} $$

Because it's trivial to derive and I've already basically laid out the whole problem for you, I'm going to leave it to the reader to have the satisfaction of this derivation.

If you start the cat in a vertical orientation, it will naturally spin as it falls to correct its orientation so as to land on its feet. Measure the height and length of your cat and determine how high you need to drop it when uncharged so that it has spun exactly ninety degrees when it hits the ground. Repeat and refine until the cat can no longer keep up - it can't spin fast enough. Be very careful here because strange effects come into play when you bring a cat to this limit.

Knowing that the cat is trying its hardest to correct its orientation, you can now charge it up and drop away - bomb bay open. Now, presuming that the cat is energized and forming a capacitor with the ground plane, the charges in its body should have separated: some to its paws and the others to the top of its furry back. As it descends, these charges will each experience a force through the magnetic field according to Lorentz' derivation above and will produce a torque on the cat's body causing it to spin relative to the mat it's on.

Continue to increase the voltage across the cat until the exerted torque matches the efforts of your furry friend to right itself. When the cat can no longer spin at all, you have all the required variables.

\$V\$ is the voltage at your final drop. \$\vec F\$ is derived from the torque on the cat based off of how quickly it was able to spin before applying the voltage. Resort to high-school level physics and your particular cat's geometry to derive this value (N.B., this need only be done once and can be saved for future tabulations). \$\vec v\$ is completely dependent upon gravity and the point in time during the fall when the measurements were made. \$\vec B\$ is the known magnetic field strength based off of the magnets you use.

If this seems too complicated for you, simply drop the cat from a sufficiently high point so that it reaches terminal velocity before starting your observations.

Finally you get the value of \$C\$ with nothing but some fidgeting, a voltage source, and keen eyes.

Conclusion

Obviously this is a simple problem most physics students have done, if indeed they have ever done real physics. The pictures are missing but it's late and I can't spend all of my time helping you out on such basic trivia. There are far more ways to do this measurement, so put your thinking cap on and let us know how it goes!