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Let's analyse the case for an infinitely long cylinder first, then move to a finite one (I've just realised you mean one of finite length having written my answer). Your hypothesis is right only in the case of an infinitely long cylinder, but not for one of finite length, unless your cylinder is a perfect conductor, in which case no symmetry assumptions are ...

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Am I correct that you can rephrase your question to 'electrons move so slow, how come that when I flip the light switch the light comes on basically instantly?'? It's true that the electrons travel very slowly. But these electrons don't have to travel across the wire to power your light bulb. In electromagnetism, we have the continuity equation $\nabla J = ... 2 When an electric dipole is placed in a uniform electric field making an angle with the direction of the field as shown in the figure. Force on charge$-q=-q\overrightarrow{E}$(opposite to$\overrightarrow{E}$) Force on charge$+q=q\overrightarrow{E}$(along$\overrightarrow{E}$) Thus, electric dipole is under the action of two equal and unlike ... 2 It seems to me that you have more of a conceptual issue than a mathematical one. To hopefully remedy this, let me remind you of a couple of facts. Given an electric field$\mathbf E$, an electric potential$V$for$\mathbf E$is any scalar function$Vfor which \begin{align} \mathbf E = -\nabla V \end{align} It follows that ifV$is such a potential, ... 1 In a conductor the electric field within itself is always equal to 0 and therefore all of the charge is found on the surfaces of the conductor, forming an equipotential. We are left with a surface charge density$\sigma$and an electric field (close to the surface): $$\vec{E}\approx \frac{\sigma}{2\varepsilon_{0}}\hat{n}$$ We are left with the natural ... 1 I've posted an answer describing the derivation of potential energy which you might want to read, as the same argument applies to electrical potential and I think that's what you're missing. Basically, given an electric field, the first step in finding the electrical potential is to pick a point$\vec{x}_0$to have$V(\vec{x}_0) = 0\$. Then, to determine the ...

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A non-zero electric field on your Gaussian surface does not mean a non-zero flux. There are positive and negative contributions to the flux due to the electric field pointing in & out in different places. Your cosine term confirms this. Apparently the contributions must cancel in this case since the net enclosed charge is zero.

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Every field line from the dipole must begin on one charge and end on the other. That means that if a field line passes out of your surface it must pass back in through it again. The surface as a whole will have the same number of field lines going in as out, so the net flux through the surface will be zero.

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Electromagnetic force is not propagated by electrons, it is propagated by photons. By definition these travel at the speed of light (in the material). Impedance and capacitance play a part in how quickly the system responds to you turning it on / connection a battery, but are generally very small in a plain wire. The electrons are moved by electromagnetism ...

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