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4

If the electric charge from lightning is captured and harnessed through circuits, eventually it will reach the ground, but once there it will join the general discharge process mentioned by Feynman. It won't lead to any build-up of charge. Incidentally, it would be very difficult to harness lighting with sufficient regularity to make a difference to the ...


3

Massless particles can carry and do carry confined charges (gluon in QCD) and they may carry charges under spontaneously broken generators (photon is transforming e.g. under the generators of $SU(2)$ associated with the W-bosons) but they cannot carry charges under unconfined $U(1)$ force like electromagnetism. The reason may be explained in different ways. ...


2

You can attract metal with static electricity. Consider the text-book example of a conducting sphere vs. a dielectric sphere in an electric field. Let's assume the field is homogeneous. This field polarizes both spheres, but in different ways: Conducting sphere: The free electrons rearrange themselves on the surface until the total electric field is ...


2

If you were holding some charge there with some force and always had then an equal charge would distribute throughout the surface of the conductor so that an equal but opposite charge could be right where you are holding your charge. So it is just like the charge was always distributed on the surface. If however you inserted some charge somewhere really ...


2

The binomial expansion says that $(1+x)^n=1+{n \choose 1}x^1+{n \choose 2}x^2 + ...$. This should be familiar to you for positive, integer n just by expanding out the parenthesis. For NEGATIVE n, it still holds, provided you interpret ${n \choose k}$ correctly for negative numbers; for our purposes, we just need to know ${n\choose 1}=n$ always. For very ...


2

Considering the charge in an enclosed surface is always 0 I'm not exactly sure what you mean by this but do understand that the Gaussian surface 'encloses' a volume of space within which the enclosed charge resides. For example, consider an isolated point charge $q$. Due the spherical symmetry, the appropriate Gaussian surface is a sphere of radius $r ...


1

I think an insulator does not completely stop charge transfer, if viewing it to act in the same way as maybe ie. a thermos cup, which does significantly increase the cooling time of a hot coffee inside, but does not fully prevent heat from escaping, hence allowing the hot coffee/material inside to cool.


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Static electricity is not like regular electricity in that it does not involve closing a complete circuit; it just needs a large difference in voltage potential between one object and another. When you shuffle your feed on a nylon carpet and touch your finger to a doorknob, you are not closing a circuit. Instead, you are building up a large negative charge ...


1

Your description is not very complete, but I guess what happens is exactly what you expected: to get significant repulsion (to counteract the atmospheric pressure), you need very high charge, which will be necessarily limited due to air discharge (maybe that is why you observed sparks). I don't see how replacing an aluminum shell with a plastic bag ...


1

Intuitive Answer: $\newcommand{\curl}[1]{\nabla \times #1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$There is an emf created because of the Lenz's Law, which means an electric field is being created. The electric field is circular (thus the force, which is proportional to the electric field, is also circular) and the charged particle would spiral ...


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Think about what exactly the rate of flow of electrons is? It is the number of electrons per second passing through! The number of electrons is not included in your expression, and that's the problem. Let start over but this time with the number of electrons $n$ included: $$Q=It \Leftrightarrow \\ en=It \Leftrightarrow\\ \frac{n}{t}=\frac{I}{e} ...


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You have to think more carefully about what exactly $Q$,$I$ and $t$ signify: In $$ Q = I t$$ $Q$ is the charge that is transported by the current $I$ during the time $t$. If you now write $$ \frac{I}{Q} = \frac{1}{t}$$ then this gives how many times a charge of $Q$ is transported by $I$ during one unit of time (second), since $t$ is the time to transport ...


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You are right! The trick to remember here is that vector fields permeate all of space (literally all of it) and field lines are only a convenient representation of this. When a new field line is added due to the increased magnitude of the field at that point in space, the field vector 'arrow' that is introduced always existed there but was just small ...


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Yes, for any pair of two different materials, it's basically guaranteed that the triboelectric effect will charge one of them positively and one of them negatively. However, for a pair of materials that are too close to each other at the triboelectric scale, the charges may be zero. See a page for more: https://en.wikipedia.org/wiki/Triboelectric_effect ...



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