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You are referring to the formula for capacitance of a uniformly charged disk. A metal disk has uniform potential, but NOT uniform charge: the charge density at the periphery of the disk is higher than at the center.

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It's not so much the increase/decrease in charge that is responsible for a change in mutual conductance, as the change in geometry, since by definition the mutual capacitance of charged conductors is independent of their charge. This answer on the meaning of mutual capacitance provides details if needed. The interesting thing though is that the effect ...

2

Because an LED acts like a diode, the negative current from the AC source will be clipped and the capacitor will always be charged and it will act like a second voltage source. Look at my picture below. You can see that the green line represents the voltage going through the capacitor and the blue line (it's a little hard to see since it's covered up by the ...

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Here is another attempt at an answer, from a slightly different angle. It is less sophisticated and may add insight. You say that the capacitors are isolated before they are connected. Literally, that is only possible if they are in different universes. In the real world, they are connected through an infinitesimal capacitance - the capacitance between the ...

2

The phenomenon you are talking about is called dielectric absorption. The way it works is this: Let's say you've just discharged a capacitor. An ideal capacitor would remain at zero volts after this. However, in real life, the capacitor will develop a small voltage from time-delayed dipole discharging (also known as dielectric relaxation). Dielectric ...

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Actually charges do accumulate even around a resistor. Consider the following model: two conductors with resistances $R_A$ and $R_B$ and lengths $L_A$ and $L_B$ are connected to each other, and a potential difference of $\Delta U$ is applied to their free ends. By Ohm's law we have: $$\Delta U_A=IR_A,$$ $$\Delta U_B=IR_B.$$ This means that, for equal-sized ...

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I think it is easy to understand. Here I am giving a heuristic picture, current in any wire is generated by charge flow. $I=\frac{dQ}{dt}$ now if you apply alternating current on the sides of capacitor you will find that charge on one plate is constantly increasing and decreasing, which induces the opposite charge on the other plate changing with same rate....

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In principle any acceleration of an electron causes some radiation, and an electron has to accelerate in order to leak from one plate to the other. However: the velocities, and therefore the accelerations, of electrons in electrical circuits are small. Calculating the electron drift velocity is an exercise routinely given to students and the results tend ...

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You obviously know the formulae for finding the equivalent capacitance for series and parallel combinations. Step back one pace to remember that capacitors in series have the same charge on them and capacitors in parallel have the same voltage across them. So solve the problem by looking at the pairs of capacitors which are in series with one another not by ...

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Let us say plate A has a charge q1 and plate B, which faces plate A has a charge q2. By making use of the fact that the net field in the bulk of a conductor in static conditions is zero, and that the net field near the outer surface of a conductor equals [local surface charge density/€0], you can prove the following: Charge on the outer surfaces of A and B ...

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Measurements disturb the double slit experiment because of the particle nature of light and matter. In order to measure which slit the electron passes through there must be some sort of interaction to detect the electron. By putting a capacitor in the way, you would drastically affect the particle. Think of it like putting a hose-pipe in front of a ...

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