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Connect the battery to this circuit. At this point, the charge doesn't know that there is a hole in the circuit. Negative charge therefor flows away from the negative battery pole, since it is repelled by this same charge, as far away as it can along the attached wire - this means that the charge will pass through the light bulb, and the light bulb will ...


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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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Both good answers. Note that the Earth has a self-capacitance by the same arguments. The Earth carries a net charge and really does have a capacitance. Remember that capacitance is defined in terms of the work that you must do to take a charge from one plate to the other. Two charged objects are involved. Even if the plates are nearly an infinite distance ...


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Let's deal with circuit elements (resistors, inductors, capacitors, batteries, meters) which have two connection ports. Two components are connected in series when they are connected to each other only once and not connected to other components at that connection point, also called a \textbf{node} (IMPORTANT term). If a third component is in series with ...


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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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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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Does it take the same amount of time to charge a capacitor as it does to discharge to get to the stable state? Yes, in an RC circuit both of these are $\infty$. will it take 5Tau to discharge a charged capacitor to its stable state like it would to charge a capacitor when the capacitor is completely uncharged? Well, that's not "completely ...


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First of all, it will never get fully (dis)charged, i.e., its voltage drop will never match the supply. The time constant is always the same, because it does not depend on the boundary conditions of your equations. To answer your question, all time scales remain the same. That means if you define the system to be stable after $5\tau$ (which is reasonable), ...


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I think you're drawing a parallel where none exists. Mutual inductance is where a changing current in one inductor influences the current in an adjacent inductor; this happens because it's easy for the (also changing) magnetic field of one inductor goes through another inductor (e.g. a transformer). Mutual capacitance, instead, is just a measure of the ...


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I have used this technique to measure the capacitance changes in a salt solution to determine how much moisture had been absorbed but it is really a hack. As the solution absorbs more water it increases the volume a very small amount and what we are really sensing is the capacitance change as a function of volume. In our instance in the amount of water ...


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You seem to be asking two questions: why is the current on either side the same, and what happens to the energy. To clarify where the energy goes, the kinetic energy is transferred entirely into electrical potential energy of the capacitor. This is the energy stored as a result of all the similar charges being close together on either capacitor plate. There ...


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The battery has to do some work in order to create a static electric field inside a capacitor.The current from the battery has to overtake the resistance in order to flow and to reach the charges into the capacitor.So some heat is generated which causes waste to energy.Here the work is done by the battery to charge the capacitor.But the force of the static ...


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The quick answer to your question is "yes". However, there may be a problem. When you ask, "Can we not use this electric field to generate current?" you have not clearly defined the amount of current you might want generated. Van de Graaff generators are not notably efficient, even among other electrostatic generators such as a Wilmshurst machine. The ...


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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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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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