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1

Why, fundamentally, are particles charged? The answer to this and the similar questions is : because that is what we have observed and defined by measurements with innumerable data. It is an existential question. Physics is not about existential questions. Physics is about observations with measurements of the way the natural world behaves and fitting ...


4

This is ultimately a very deep question. We do not have a very easy ability to answer it at this time. Let me give you the pieces that we presently have. The basic interactions. The world as we know it today consists of five fundamental things that happen; they are all called "fields" and the first four are called "forces" or "interactions". In rough order ...


1

The standard model of particle physics is built on gauge theory, which is a theory of local symmetries. This means that the symmetries act differently at different points in space (and time). For example, if you move everything in the universe 1m in one direction, nothing is changed. This is a global symmetry, because everything has been moved by the same ...


4

I am not sure how to answer all of your question but I can answer the neutron and proton part. The charge difference is due to their composition. A neutron is composed of two down quarks and one up quark. A proton is composed of two up quarks and one down. Up quarks have a charge of +2/3 and down quarks have a charge of -1/3. A neutron with composition udd ...


1

Your work is fine (depending on your units) and what you were asked to show is wrong. Though I do object to saying you have a force equal to $ma_c,$ I would just say that a net force orthogonal to the velocity makes it go in a circle of radius $r$ where $F=mv^2/r.$ And the problem is famous. The fact that the frequency doesn't depend on the velocity or the ...


0

What happens if the particle is on the circular path? Hint. There must be something compensating the Lorentz force. EDIT: I think your identity for the angular velocity is wrong anyways. It should be $$\omega = \frac{qB}{m}$$


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

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 ...


0

A conductor can have a zero net charge. One way to do this is to have an inner surface with a net charge of $-Q$ and an outer surface with a net charge of $+Q.$ So for instance you can have a surface charge density of $-Q/(4\pi R^2)$ on the surface $r=R$ and you can have a surface charge density of $+Q/(4\pi (2 R)^2)$ on the surface $r=2R$ and this has zero ...


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 ...


0

The fact is that metals do get polarized but in a different way. Since electrons are rather free, they distribute themselves along the frontier of the system, being attracted by the field. But they do so in such a way that the field becomes zero inside the metal, because the contribution from the external field and these electrons cancel each other. You ...


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

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

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.


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 ...


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. ...


0

When you connect the negative terminal to ground, why would the negative charge flow anywhere? It has to exist somewhere, and it might as well be close to the positive charge on the other capacitor plate. You could also consider a very similar scenario in which you connect the negative side to ground before disconnecting the charging source. In this case, ...


1

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 ...


0

But according to the given answer if the electric field $E$ is changing then the flux should also change since $ds$ is constant. As you say, $E$ does change. However, it changes both where it enters $S$ and where it exits $S$. In the integral, the sign of $E \cdot ds$ depends on whether the field is pointing into or out of the surface. To make this more ...


0

In regard to the first, we have a 5v supply, which from my understanding means that if you were able to enclose a coulomb of charge eminating from the negative terminal, you'd find that it has 5 joules of energy. This isn't the typical understanding. A 5V (ideal) supply (source) maintains a 5V potential difference across the terminals independent ...


0

First, a couple of asides: A. When circuits are drawn, the convention is that straight lines represent perfect conductors. Any real resistance is shown as a resistor, and hence, gets included in any IR drops. You don't go and say "well, there are some other resistances." If they are important, they are shown. Circuit analysis of diagrams is done with this ...


0

I'd like to start with a slightly pedantic aside: I think you should try to avoid the phrasing of electrons "having energy"; what they have is potential energy, and potential energy is relative. This is important to remember. Question 2 is trivial, so I'll answer that first: there is no such thing as zero resistance, so the wire from R3 to the batter's ...


0

Power in a circuit is defined by P= I x E P is power in watts or joules per second I is the flow of electrons in coulombs per second or amps E is the electormotive force or voltage the energy of the electrons is lost to heat your first question the power supply see only the circuit as a whole it dose not see the individual resistors for R1 the total ...


0

By Gauss's law, the scenario you describe can only happen when there are electric charges within this volume. If they are positive charges, you need to draw field lines coming out of them. If they are negative charges, you need to draw field lines ending at them.


3

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 ...


-1

I don't think this situation is possible.Consider finite field lines passing through a given area vector in vaccum.Now If you enclose the given field lines and charge by Gaussian surface where one of the plane is parallel to x-y plane.By Gauss's law the net electric flux is constant. While changing the magnitude of electric field with same enclosure i.e ...



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