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The voltage becomes the same as Earth, but this doesn't mean that the charge goes to "zero". By "zero" here, I mean that the positive charges (nuclei) are perfectly balanced by the negative charges (electrons). You can call the voltage of Earth 0 Volts, but this is a relative measure. Charge, in the usage here, is not a relative measure because it is a ...

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Indeed, the $\vec{E}$ field in a parallel plate is independent of distance from the plate. This works because of the assumption $d \ll$ length of plate (thus, we can ignore side effects of the plate). And as Bort pointed out, it is the Voltage $V$ that scales linearly with respect to distance from the plate, while $\vec{E}$ will remain constant.

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All you need to charge a battery from a capacitor is to have more voltage charged on the capacitor than the voltage of the battery. The size will only affect how much time the capacitor will charge the battery. If you could charge the capacitor over and over and discharge it into the battery every time it was full it would eventually fully charge the ...

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I'm assuming $V_R$ is taken to be the voltage across the resistor in a series RC circuit. The transfer function comes directly for the voltage division rule: $$\frac{V_R}{V_\text{in}} = \frac{Z_{R}}{Z_\text{series}} = \frac{R}{R+\frac{1}{i\omega C}} \, .$$ In this equation $V_R$ and $V_\text{in}$ are phasors, meaning that the actual time dependent ...

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I stumbled upon this similar question. This OP is more general and I found the way to extend the answer. Apparently the plates are coupled to the environment through a capacitance as well. By superposition we have $Q_1 = Q_{12} + Q_{10},$ $Q_2 = -Q_{12} + Q_{20},$ $Q = Q_0 = -Q_{10} - Q_{20},$ where the index $i = 0, 1,2$ corresponds to the $i$th plate ...

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