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First point is that you would normally get more charge at the edges. Second point - if you increase the gap then the capacitance would be reduce and the quantity of charge that could be stored for a particular potential difference would be reduced. Remember electric field is charge / distance so decreasing the gap would increase the electric field You ...

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Let's do some calculus. Suppose you have two plates, almost parallel (off by an angle $\alpha$). The plates lie in the XY plane, from $(0, 0)$ to $(x_1, y_1)$. At $x = 0$, the plates are separated by a distance $z_0$, and at $x = x_1$, the plates are separated by a distance $z_1$. We'll now consider an infinitesimally small element of both plates. (Since ...

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It helps to think about Maxwell's equations here. The first one of which is $$\nabla \cdot \mathbf{E} = \rho / \epsilon_{0}$$ where $\rho$ is the charge density. Electrostatics is always like a chicken and egg problem: charges make fields, but fields move charges around, so it can lead to some confusion. It turns out that in the steady state ...

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The "derivation" you describe is valid at a particular moment in time. \begin{align}\Delta E &= V \Delta Q\\ \frac{dE}{dt} &= V \frac{dQ}{dt} \\&= VI\\ P(t) &= V(t) I(t)\end{align} I added the dependence on time explicitly. To address your comments: $W=QV$ is only true when $V$ is constant; you can't simply take the derivative and ...

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Power is an instantaneous concept. $P=IV$ gives the instantaneous power at a given instant of time, given $V$ and $I$ at that time.

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I found the solution to my problem : The equations can be written : $$I = V \times C_{ij} \times i\omega$$ Where $I$ is the vector of known currents going threw the electrodes, $V$ the potential of the electrodes and $\omega$ the rotational frequency of the current. In this case the potential of the transmitting electrodes is known and the current ...

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Can we just calculate by using 1/2 CV^2? Yes. If $v_C(t)$ is the instantaneous voltage across the capacitor, the instantaneous stored energy is just $$U_C(t) = \frac{C}{2}v^2_C(t)$$

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The equation for capacitance is Q=CV or V=1CQ. I don't understand what is the physical meaning of this "C": Does the charge in a system changes linearly with voltage under all circumstances? This first part is the statement of the behavior of an "ideal" capacitor. Because it is an idealization, it is easy to characterize as a linear component whose ...

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Worth noticing the difference between capacity and capacitance. You want the latter to have the meaning of the former - when instead it just describes how much the voltage increases when you add charge. It doesn't tell us when the capacitor is "full" - for that you need to know the rated voltage as well as the capacitance. When a capacitor is used in a ...

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In answer 1) Yes, charge will flow to make the potential of the wire the same as the + terminal of the battery given that the wire is neutral or at earth prior to making contact. The ammount of charge that would flow would depends on the difference in potential between the wire before you attach it to the battery and the battery - we could calculate that ...

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The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a ...

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This is a nice example of one of the foundational issues in relativity: how do we know that energy-momentum transforms like a four-vector, or, essentially, how do we know that $E=mc^2$? A historical overview is given in [Ohanian 2008] and [Ohanian 2009]. As Ohanian points out, there are logical problems if one tries to do what Einstein did in 1905 and prove ...

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To answer this question, one must reconcile three apparently irreconcilable items: The boosted electric field between the capacitor plates is unchanged (since it is parallel to the boost). The boosted capacitor plate separation contracts. As the time-element of a 4-vector, the potential difference across the capacitor should increase when boosted. The ...

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Let us first consider a capacitor that is charged and not connected to a battery or other electric power source. I think you need to take into account that the Electric field in the capacitor is reduced by inserting the dielectric and also the voltage drops between the plates by inserting the capacitor. The dielectric is pulled in to the cap as the ...

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Why is it so? Well, it isn't actually always so. It will depend on the actual circuit configuration and whether the switch opens or closes when $t=0$. But first, here are a couple of crucial results to always keep in mind when solving these type of switched circuits: the current through an inductor must be continuous the voltage across a capacitor ...

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@Floris' answer being good, i'll give another view on the matter. A capacitor is equivalent to an open circuit (since simply put, a capacitor is an element consisting of two plates which do not actually touch but through another medium, the dielectric, the circuit is not connected at that point where the capacitor is located), whereas an inductor is ...

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Circuits with inductors are sensitive to changes in the signal - think of them as differentiators. Circuits with capacitors are responding to the integral of the signal over time. When you first turn on a circuit, the current wants to make a step change - which the capacitor doesn't care about, but the inductor resists vigorously. Thus the current will flow ...

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