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9

The standard model is very successful in its group structure in ordering all observed particles. To introduce a particle with charge and zero spin, you will need a different model that would also accommodate the symmetries observed experimentally and fitted by the standard model. So the answer to "why" is "because" we have not seen any and can model well ...


5

It's a matter of history. When George Stoney developed Stoney units in 1881, or when Robert Millikan performed the oil drop experiment in 1909, it wasn't yet known that it was possible for anything to have a charge smaller in magnitude than the charge of an electron. By the time the quark model was proposed, in 1964, the use of the "elementary charge" being ...


3

Well discrete charges, and in particular point charges, are a consequence of quantum mechanics. If you're considering just the classical theory there are no special conditions on the charge distribution. I'm not sure I'd say charge density was more fundamental than charge, but charge density would be what gives you the divergence of the electric field.


3

It will help you understand the quantum mechanical picture if you read up on atomic orbitals. These are the loci around the nucleus where the electrons have a probability to be found. You will see that the orbitals have a shape, which depends on the angular momentum of the state. The electrons carry the charge and thus you can interpret the plots as ...


3

The Lagrangian $$\mathcal L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \mathcal L_\textrm{free} + eA_\mu J^\mu \tag{1}$$ where $A_\mu$ is the 4-potential, $F_{\mu\nu} = \partial_{[\nu}A_{\mu]}$ is the field tensor, $\mathcal L_\textrm{free}$ describes fields other than $A_\mu$, and $J^\mu$ is the 4-current density expressed in these other fields, describes a ...


3

Using cylindrical coordinates with the origin at the center and the $\phi = 0$ direction 'down' (the OP says the image should be rotated CCW 90 degrees), the electric field appears be have only a radial component with a sign change for $\phi = \frac{-\pi}{2}$ and $\phi = \frac{\pi}{2}$ $$\vec E = E(\rho,\phi)\hat\rho $$ $$E(\rho,\phi) = ...


2

Yes indeed. To get the result you stated from Gauss' Law you must asume that the charge is distributed in such a way that you have a spherically symetric field. How you get to that field doesnt matter though. So the charge might all be concentrated at the center or all lying on the surface of a sphere: it doesnt matter as long as the field is spherically ...


2

The electron charge is called $-e$; let me pick the convention where $e$ is positive. The atoms (e.g. the hydrogen-1 atom) contain the same number of protons as electrons and they are neutral. They must be neutral because ordinary macroscopic matter is composed of atoms and it has to be neutral because the atoms would otherwise attract the oppositely ...


2

There is no such thing as perfectly non-conducting. We just simplify things into conductors and insulators. A current will flow from the plastic comb through you to ground, just not as quickly as it would from a metal object connected by a copper wire. The comb can pick up paper because the paper isn't a good conductor. The electric field from the charged ...


2

In classical mechanics, it is often possible and convenient to describe a system with an object called a Lagrangian (in that it governs a system's behaviour, the Lagrangian is similar to a Hamiltonian). Like the Hamiltonian, the Lagrangian ought to be real - and any terms inside the Lagrangian ought to be Hermitian. In quantum field theory (QFT), the ...


1

If you have a complete circuit, every piece of metal will gain and lose the same number of electrons and will not have a net charge. If you connect two plates, one to each end of a battery, the battery will take charges from the plate connected to the positive terminal and send charges to the plate connected to the negative terminal until the voltage ...


1

A charged conducting material in the form of a sphere or an infinite plane can only be uniformly charged in the absence of external charges. Any other shape of charged conducting material can be induced to be uniformly charged by the placing the right external charge density around it. A conducting infinite cylinder is also uniformly charged (in the absence ...


1

As long as your product $|q_{1}||q_{2}|$ remains the same as in the case where you had the equally charged spheres, then yes, you will get the same value for the angle (provided the masses are equal). This is because the electrostatic force acts equally on both charged spheres.


1

I've found an explanation. As Duncan said, as the sum of the charges stays equal the greatest product of two integers with a certain sum is half of the sum when (for example 5+5=10 and 5x5=25 is the greatest product. NOTE: Here the sum is always even as we take two equal charges 'q' initially). Hence all other series yields a force lesser than the ideal ...


1

Voltage has absolutely nothing to do with charge. I can "move" an infinite amount of charge trough a superconductor with zero voltage. Are you asking about the relationship of charge to voltage on a capacitor? That's a linear relationship: Q=C*U. The charges, in that case, are not "created" but merely separated. If you want more charge for the same amount of ...


1

It is very hard to use a voltage source to induce charge on an insulator. The reason is that by definition, an insulator does not conduct electricity - so if you apply an electrode at one place, you will not move electrons elsewhere, and so you cannot induce a net charge (the best you can hope for is to create polarization, and maybe pull off a handful of ...


1

The Triboelectric effect is the process through which materials can become electrically charged through friction when they come in contact with other different materials. These materials do not have to be insulators for this effect to take place however if they are good conductors the charge will usually flow away. There is a series of materials ranging ...


1

When you put a charged insulator in air, the reason you lose charge is mostly due to humidity in the air. I gave details of this mechanism in a recent answer to a related question: http://physics.stackexchange.com/a/130988/26969 The curves in the referenced paper (some of which I reproduce in that answer) show you how the leakage current is a function of ...


1

Wikipedia> Electric charge: Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. Electric flux: In electromagnetism, electric flux is the rate of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a ...


1

As the comments say, it would do nothing if you change all negative charges to positive and viceversa for all know particles. You actually have a real physical example: antimatter (which in most theories behaves just as standard matter but there might be some non-symmetries when you include all particles (it depend on the specific modification of the ...


1

The Higgs is part of a complex scalar doublet in the standard model. It. carries both hypercharge and weak charge. So we have discovered charged scalars. Now perhaps you are only interested in ELECTRIC charge. So does the Higgs doublet carry this? Well once the Higgs picks up a vev, then some parts do and some parts don't. The parts that do carry electric ...


1

You can think of light as the carrier of the electromagnetic interaction. The particles interact with light, not directly with each other. It is an experimental fact that light does not interact with itself. Note that this is not the case with quantum chromodynamics (the theory of nuclear matter). This theory is built along the same lines as quantum ...


1

Before we talk about the term "spherical shell" and "thin sphere", let us talk about the possible cases of the sphere itself. Generally speaking, it depends on what is given to you. If you have been given $\rho _s$, then probably it will be hallow. If you have been given $\rho _v$, then definitely it will be solid. Keep in mind that whether it was ...


1

I have came up with this: Charges are the sources of the electric field. So, whatever the point that field lines are "created" or "destroyed", must be a charge. Then, if there are a charge, then must be on the center. Calculating the electric flux: $$ \phi = \iint_S\ \mathbf E\cdot d\mathbf s = \frac{Q}{\epsilon_0} $$ Let's pick a sphere as gaussian ...



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