# Tag Info

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The fact that the static charge does not spread uniformly is the basis for things like lightning rods. Sharp edges are places that static charges, particularly higher voltage ones, like to reside. This design also aids in dissipating dangerous voltages via the coronal (ionized air) discharge mechanism. Sometimes, the uneven charge distribution is because ...

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Yes, the electric charge, of the quark, was truly made before $\Delta^{++}$. One found the electric charge of the quark first. Then, second, it was named $\Delta^{++}$. The new delta soon formed answers of new electric charges to other elements. Then formed a new omega, dew to the other new electric charge.

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It depends. Is the electric current flowing briefly in a circuit that contains a large capacitor? None of the other answers seem to have contemplated this possibility. If it is a steady current through a resistor, definitely the answer is no.

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When charging a metal object with induction, another charged object is brought near to that neutral metal object but it is not touched. This will make the electrons move within the metal object according to polarity of the charged object. Keep in mind that opposite charges attract one another whereas like charges repel. This movement of electrons creates an ...

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Your question is very good. The answer is very easy, but very deep though. "Charge $Q$" means that there is a conserved (quantum or classical) number $Q$ respect to some symmetry $G$. i.e. the system is invariant respect to certain symmetry $G$ transformation. (You can derive it from Noether theorem or simply the transformation on the "fields" in general, ...

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Although theories of electromagnetism (a la Maxwell) and electroweak unification (a la Weinberg, others) work well to explain the behavior of charge / magnetism / electroweak, none of these provide a mechanism for originating electric charge. Similarly, in QCD, which explains the behavior of the strong force between quarks and gluons, no mechanism is ...

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Jerk_dadt is correct. Electric current is the flow of free electrons in the conductor. At any instant, the number of electrons leaving the wire is always equal to the number of electrons flowing from the battery into it. Hence, the net charge on the wire is zero. If you say the current carrying conductor is charged, it will violate the Kirchoff junction ...

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No, not necessarily. Current is the just the movement of electrons already in the wire (that is neutral). The electric field (or voltage) applied causes them to have a net movement in one particular direction (i.e. opposite direction of conventional current). So copper has 29 protons and 29 electrons per atom. A copper wire would have a net zero charge. It ...

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Like many things in electrostatics, this class of problem will come out with an unintuitive answer. Would the liquid touch the object? The thinking of the question is that the fluid is positively charged, so the fluid is attracted to the object. For the bulk fluid, this would be true. By that logic, it would be somewhat safe to say that the electric ...

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This is a interesting question. This is how I reasoned it out. I think there are three situations here (assuming rotationally invariant): 1) The object is very deep where it cannot overcome the electrostatic force from all the positive charge above it, with only its Buoyant force. 2) The object is close enough to surface to where the buoyant force and the ...

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electrons will only distribute evenly if the conductor is spherical. if it is not, coronal discharge will occur at sharp points until the voltage is too low to sustain it. of course, breakdown will occur even with a spherical conductor, e.g., van de graff generator, but much more charge will accumulate before any one point has sufficient potential to cause ...

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You are basically confining two negative charged small spheres in a bigger, neutral sphere (this image is better since the you don't need to worry about quantum effects, and there's no need to consider unphysical situations). Since both spheres would repel each other, they'd try to maximize the distance between them, so they would set up in a configuration ...

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I was wondering why, in the derivation, they set charge to Q... but if the definition is amount of charge being able to be store $Q$ is the charge separated, not stored. A charged capacitor is not electrically charged any more than a charged battery is. Rather, a charged capacitor has stored energy. To charge a parallel plate capacitor, electric ...

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By charge on capacitor, we (generally) mean the magnitude of charge on one of the plates.

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You have to be careful of your Gaussian surface. If you include both charges (from both plates) then your Gaussian surface is outside of the parallel plate capacitor and the electric field is indeed zero because there is zero net charge when encapsulating the parallel plates with your Gaussian surface. However, if you place one part side of your Gaussian ...

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An experimentalist's view: I do not see the need to search further for why the three quarks add up to the electron charge than that given by the group structure of the Standard Model. The SM is very successful in organizing into beautiful symmetries the particle and resonances data gathered the last sixty years or so. There is no experimental reason to ...

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We have to make some assumptions. Let's first consider what we mean by a neutral conductor with small balls of negative charge representing electrons. For the positive charge, let's assume a continuous uniform background ("jellium", if you will) whose total charge is equal to the total charge of the electrons. Let's also assume that the electron balls are ...

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if electrons were balls, there wouldn't be conductors. Coulomb systems are unstable. you wouldn't be able to have free electrons and other stuff, they'd all lump together with positive nuclei. the remaining balls would run away as far as the Coulomb force pushes them.

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Also, if a charged, insulated wire with Direct Current applied tension exhibits coronal dischage over a period of time, the insulator must be conducting and shorting to the outside. Alternatiing applied tension is much messier to analyze, but involves the dielecctric constants and the total capacitances of both the insulator and of the surrounding air. ...

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. . . Well, anyway, you can get a shock through the insulator when it is under tension from a power supply or source and the tension exceeds the breakdown threshold of the insulator. The conductor inside the insulator shorts through it and can shock you. The other case, with no breakdown, depends on whether or not the dielectric in the insulator ...

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Your answer is correct. Indeed there is surface charge deposited on the surface of the insulator. This configuration is called a Dielectric barrier discharge. Its concept of work is explained in this document. In section 2 it is clearly stated that the charge deposited on the surface of the dielectric play an important role in sustaining the discharge. The ...

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Yes it does. Think of the voltage as an electric field and hence the stronger the electric field (and similarly larger voltage) the stronger it will push charges. Consequently, even more charge will accumulate on both leaves (of the same charge) and hence there will be a stronger "Coulombic" repulsive force between the two leaves, and hence farther ...

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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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It should be associated with the work function, which is the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface, and different materials have different work functions. Consider a very simple case, that a spherical electrical object exists in vacuum. Considering ...

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Electric charge did not exist before the Higgs field acquired its vacuum expectation value. It was only for a very short time just after the big bang. At that time there was weak hypercharge and weak isospin. Only after the Higgs field was turned "on", these 2 charges started to mix and thus electric charge was born. Before that there was also no ...

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From the Guass law it follows that for a solid spherical shell, charge on the surface also behaves as if concentrated at its center. So, instead of spreading $+Q$ charge on the surface and placing $+Q$ charge at the center, you can directly spread $+2Q$ charge on the surface for the same effect. So, considering this, yes it is possible to create such a ...

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If the sphere is conducting, then the field at any point within the conductor is zero: the field within any ideal conductor is everywhere zero. So if you displace the center charge, it will just stay where you put it. It will not move to the surface. Consider the same situation but with vacuum inside the sphere instead of an ideal conductor. The field ...

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As you say, the electric charge is quantized and the charge on a charged body is always an integral multiple of the charge on a single electron or a proton. In most practical situations, charge on a charged body is large as compared to the magnitude of charge on an electron or a proton that the quantization of charge may be ignored. In other words, we can ...

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Hint : Speed of particle remains constant as magnetic field is always perpendicular to it. Also try to find a relation between the theta , D and h. You can also include R if you want. Try to draw the circle and its centre by using that the fact that normal at a point of a circle passes through the centre. Feel free to leave a comment below if you have any ...

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The charge of a particle is completely independent of its mass. If you had some technique (and there isn't one) to just remove charge but keep the proton exactly the same it would not change its mass. Particle masses arise from a combination of the mass of its constituents and their interactions (the potential energy of particle interactions give it mass ...

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The potential energy of a particle with charge $q$ in a conservative electric field $\vec E$ is $$U = q\phi$$ where the electric potential $\phi$ is related the electric field by $$\vec E = -\nabla \phi$$ Thus, the electric potential is defined, up to a non-physical constant, by the associated electric field - no test charge enters the picture. The ...

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The value of electric potential difference (aka voltage) is independent of the choice of test charge, including its sign. When determining the electric potential difference between two points, you can imagine either a positive or negative test charge, which ever one tickles your fancy. Note that the same cannot be said about electric potential energy. As a ...

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I think you're confusing "electric potential" and "electric potential energy". I don't blame you, it's an unfortunate bit of terminology. The electric potential energy is defined for a particular charge (or distribution of charges), and so you need to explicitly put in the charge of the point or distribution of interest. That is, we don't talk about the ...

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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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The information about beginning of the flow of current is transmitted through the propagation of electromagnetic waves and not with drift velocity of electrons. Hence, any electric appliance turns on almost instantly, when the switch is closed.

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Although the electron velocity is very low, which is propagated almost instantaneously is the electric field. This causes the effect that all the electrons in the wire to start moving simultaneously (almost).

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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 = ... 0 So I know that the drift speed of electrons is usually pretty slow. Yes, if$10A$of current is maintained in a conductor of cross-section$10^{-4}m^2$, with number density of electrons equal to$9X10^{28}m^{-3}$, drift velocity of free electrons will be$0.000006ms^{-1}$(with the centimeter scale in your geometry box, it will be$0.0006cms^{-1}$). ... 7 When it comes to fundamental charges, the (left-handed) up-type quarks actually have either the same values of the charge as the down-type quarks, or exactly the opposite ones. It just happens that the electric charge isn't a fundamental charge in this sense. Let me be more specific. All the quarks carry a color – red, green, or blue – the charge of the ... 0 If you are looking for symmetry, I think one should point out that there IS a particle with a -2e/3 charge and a particle with a +e/3 charge. They are the up antiquark and down antiquark respectively. Now, following that, you would very reasonably ask the question why we observe more up quarks than up antiquarks, and other follow-up questions like ... 0 What you have done is calculating the densities for which the net charge density is zero. That does not mean the field is zero. You may be confused with the the principle that there is no field inside a conductor (and hence no net charge). This is not the case here. What the hint implies is that you need to calculate the field each object produces inside ... 7 The names up and down don't refer to electric charge$Q$but are rather references to isospin charge$I_3$. 6 A neutron consists of three quarks$u d d$(up down down quarks). The up quark(u) carries charge$2e/3$and the down quark(d) carries a charge$ -e/3 $. Thus$2e/3-e/3-e/3=0$18 The up quark has a charge of$+2/3$, the down has a charge of$-1/3$. If you have a bound state of charged particles, the total charge is just the charge of the elementary constituents. The neutron consists of one up quark and two down quarks, so the total charge$Q$is: $$Q = 2/3 + 2 \times (-1/3) = 0$$ 3 Because$2/3-1/3-1/3=0 $. 1 If there is an electric field, then there will be a force on a charged particle since$\vec{F} = q \vec{E}$. If you put a single charged particle in perfect vacuum, then if you add another charged particle then both particles must either move away from or towards one another, depending on whether the charges of the two particles have the same or opposite ... 2 Multispecies spinor QED is described by the following Lagrangian: $${\cal L} = \sum_i\bar{\psi} _i i \left( \partial _\mu + i e _i A _\mu \right) \gamma ^\mu \psi _i - m \bar{\psi} _i \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu }$$ If$ e _i $is the same for every particle then we have a flavor symmetry under,$ \psi ...

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Suppose such a particle existed. Question is what would happen if it was to enter an electric field? Consider $p$ ($m = 0$, $q > 0$) entering an electric field $E_i$, on a manifold $M (i,j)$ $$F_i = q E_i \; \; \;\text{but} \; \; \; F_i = m a_i$$ It follows that $F_i = 0$ since $m = 0$ meaning either $q = 0$ or $E = 0$, but such is not the case, $F_i$ ...

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We have long been taught that electric charges are neither created nor destroyed. No, we have not been taught that. We've been taught that electric charge, i.e., the net electric charge, is conserved. Imagine that, within some volume there is some net electric charge Q. Assuming there is no current through the boundary of the volume, we are taught ...

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