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Let's think about the circuit you drew. It contains a battery, switch, bulb, and a very large inductor. In fact, the inductance of a wire that goes 10 times around the Earth can be calculated (I am going to assume an air core - in fact there is a piece of iron in the middle of the Earth which makes the resulting inductance greater). $$L\approx N^2 R \mu_0 ... 2 E_1 = E_2 . since E is independent of dielectric as long as potential b/w plates is constant.$$E= = -\frac{dV}{dr}$$So, it is independent of dielectric b/w it. So, correct statement would be$$E_1d = E_2dEd = Ed$$2 Okay, draw a circle around the spherical void; done? Then see is there any charge? If not, there should not be any divergence of electric field, know that? So, from Gauss Law,$$\text{div} \mathbf{E} \cdot dv= 0$$which implies$$\mathbf{E}\cdot d\mathbf S= 0 \implies \mathbf E= 0 over that volume . Edit: Don't know what OP is trying to convey ...

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I like to think about potentials as a hilly landscape. A ball rolling in this landscape will have an energy relative to how high the hill is wherever the ball is at. If it is up higher on the hill, the ball will have less kinetic energy, and lower in the valley, the ball will have more kinetic energy. The amount of kinetic energy the ball has is related ...

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Capacitance is about stored charge - more electrons flowing into something than flow out. This can happen in a piece of wire, although it can take a large amount of applied voltage to accumulate a small amount of excess electrons. In other words, a simple piece of wire has very low capacitance. Even a straight piece of wire will have inductance because any ...

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In general, no: the field of a charge distribution $\rho$ is not the same as the field of a point charge at some point therein, except for some very particular cases (the one that everyone should know is that any spherical shell of charge has an inner field of 0 but an outer field that looks exactly like all the charge is located at the center point. A ...

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You appear to have some impression of the electric field which has given you the idea that it's some sort of bubble or something around an electron, for example. It's not. There's no real way to answer your question other than that that I can think of. It doesn't somehow fit itself through the slit or anything like that really.

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When you first connect the source, there is a very brief transient during which the steady-state DC solution is set up. The speed of the signal, i.e. the electromagnetic wave front that carries the information along the wire, is a bit less than the speed of light because of transmission line effects. Figuring out exactly how long the transient lasts would ...

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As long as the DC component does not saturate the core of the transformer, the (lower frequency) components of the waveform should be induced in the secondary. Consider, for example, the output transformer of a single ended class A triode audio amplifier Image credit In this case, the primary current is 'pulsating' DC, i.e., the primary current varies ...

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Yes surely, The pulsating DC is impure dc. Each pulse will be creating a change in magnetic flux in the transformer core. If you see the normal ac diagram the wave from 0 to T, It is similar to your pulsating DC diagram, there is a change in flux in transformer in this case. But it interesting to note that the transformer will give the increased or ...

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It depends exactly what you want this to do. The nearest thing might be a high frequency discharge from a Tesla Coil channeled through an Argon gas stream

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You seems to assume both capacitors has the same plate separation $d$. So, lets assume that. Assume there is no dielectric material. Therefore, nicely $Ed = Ed$ in both capacitors. Which is nice. :). Now, I think I understand your confusion. Have an isolated capacitor with electric field inside plates of $E$. Insert dielectric $K$. Under this case, the ...

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