4

Gary Godfrey beat me to the spherical cow joke. For a slightly more sophisticated theorist's answer, let's consider an ellipsoidal cow instead. According to the Digital Library of Mathematical Functions, the inverse of the capacitance of an conducting ellipsoid with semi-major axes $a$, $b$, and $c$ is $\frac{1}{C} = R_F(a^2, b^2, c^2) / (4 \pi \epsilon_0)$...


4

Instead of the old physics joke of "consider a spherical cow", let's consider a spherical spoon of radius r. The capacitance of a sphere in outer space is \begin{align} C_\text{sphere} &= 4\pi \epsilon_0 r=4\pi\times 8.8\times10^{-12} \frac{\rm F}{\rm m} \times r \\ &=111.\times 10^{-12} \frac{\rm F}{\rm m} \times r \\ &\approx 1~\...


3

$ \frac{9dQ^{2}}{10\varepsilon_{0}A} $, and I claim that all of this energy would be wasted on the resistor. This is where you go wrong. Some electrostatic potential energy will remain in the circuit. The only way there would be no remaining potential energy is if the voltages on both capacitors was zero after a long time. But that cannot happen due to the ...


2

My question is given that a capacitor creates a two charged sides, by the electrons jumping from one plate to another thereby making one having an excess of electrons making it negative and the other positive. If you mean by "electrons jumping from one plate to another" that the electrons move across the space between the plates, that is not the ...


2

Use an oscilloscope. Let one channel display the voltage across the capacitor. Let a second channel display the voltage across a resistor, which you have wired in series with the capacitor.


1

I have never seen a Hamiltonian for a circuit with a diode, and I doubt that it exists or used - for the reasons I describe below. However, from purely academic perspective it is an interesting question to ponder. Unlike inductance and capacitance, which can be characterized by linear response and therefore described by quadratic Hamiltonians, a diode is a ...


1

This is a nice question, Consider this circuit first, that build from one resistor and battery, We can apply kirchhoff's law and we can get that $$V_R=V$$ while $V$ is the battery voltage, thus we can say that the resistor is connected parallel to the battery. Now check the following circuit, We can see that applying kirchoff's law here will yield $$V= V_{...


1

When the separation distance $d$ increases, all other things being equal, the capacitance is less, as you already know. But the charge doesn't "go anywhere". You don't "lose charge". With no voltage applied to the capacitor, the charge on the metal plates consists of an equal number of mobile free electrons and protons. The total charge ...


1

The capacitor does not know, what changes the dielectric behavior, but in usual circumstances only water vapor comes in the capacitor. Probably it would not work if you expose it to other vapors, but how should this happen? If it shall measure the relative humidity and not the absolute it has to be coupled with a temperature sensor.


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