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Let me add two references to points already mentioned in this discussion: Today, there is no reason known why the electric charge has to be quantized. It is true that the quantization follows from the existence of magnetic monopoles and the consistency of the quantized electromagnetic field, which was shown first by Dirac, you'll find a very nice exposition ...

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The maximum charge a capacitor stores depends on the voltage $V_0$ you've used to charge it according to the formula: $$Q_0=CV_0$$ However, a real capacitor will only work for voltages up to the breakdown voltage of the dielectric medium in the capacitor. So in reality, for every capacitor there is a maximum possible charge $Q_{max}$ given by: $$... 9 It's not a mistake, and conventional current is not wrong or backwards. The labeling of one polarity of charge as "positive" and the other as "negative" is totally arbitrary. It could be done either way and everything would still work out the same. Franklin didn't choose wrong; he just chose. Labeling protons as negative and electrons as positive wouldn't ... 9 It seems you are contrasting the speed of propagation of current with the speed of the individual charge carriers. These two things are clearly separate. There are many examples. Consider sound. A fire cracker goes off at the other end of a football field from you. You hear the sound a few 100 ms later. The air molecules that were by the firecracker ... 8 Dear asmailer, the reason is simple and completely understood: the electric charge is the generator of a U(1) symmetry which is compact and may be parameterized by an angle, \phi. So wave functions may only depend on the angle \phi in a periodic way, \exp(iQ\phi) where Q is integer (or an integer multiple of e/3, if I look at the elementary ... 8 Charge comes from discrete symmetries and is countable and additive. Mass comes from continuous 4d space, is exchangeable with energy and, in quantum mechanical dimensions not linearly additive, thus not countable. Suppose you have an elementary quantum of mass, m_q. In the world we know two such quanta would not end up as 2m_q. One would add the ... 7 Actually, mass and charge are only superficially similar. Yes, they both appear in inverse square force laws, namely Newton's law of gravitation and Coulomb's law of electrostatic force, but both of those are approximations. Coulomb's law ignores quantum effects, which is a very slight approximation, but Newton's law ignores all of relativity, which makes a ... 7 Coulomb repulsion it is. Specifically, if a black hole has a lot of charge, then particles with a high charge-to-mass ratio will be repelled. Anything that falls in will contribute "more mass than charge," heuristically, keeping the charge-to-mass ratio of the black hole from getting too big. 6 In quantum field theory and its extensions including string theory, the electric charge is a generator of a U(1) symmetry which should be promoted to a local symmetry i.e. gauge symmetry. In string theory, the U(1) symmetry and the gauge field often appear as parts of the low-energy effective action. This could be enough to answer the question: we ... 6 There is a limit on how much charge a black hole may have: http://en.wikipedia.org/wiki/Extremal_black_hole In general, rotating, charged black holes is described by a Kerr-Newman metric. Intuitively, eventually the Coulumb repulsion is enough that a charged particle which does not contribute more mass than charge will be repelled. 5 There is a way of seeing this more explicitly with the Reissner-Nordstrom (RN) metric$$ ds^2~=~-F(r)dt^2~+~F(r)^{-1}dr^1~+~r^2d\Omega^2 $$where the F(r)~=~1~-~r_0/r~+~(Q/r)^2, r_0~=~2GM and Q the charge in length units. The metric has two critical points$$ r_\pm~=~\frac{r_0}{2}~\pm~\frac{r_0}{2}\sqrt{\frac{4Q^2}{r_0^2}} $$These are the outer ... 5 Charge discreteness is the statement that charge comes in packets which are of size 1 electron charge. You can understand this as a consequence of the fact that everything is made of particles with definite charge, but then it leaves the question of why the antiproton and the electron have the same charge. All particles have opposite charge to their ... 5 General Relativity is a mathematical model that relates the curvature of spacetime to an object called the stress-energy tensor. In many cases the stress-energy tensor is dominated by mass and you can simply consider the curvature as being related to the mass. However this isn't always true as I'll mention below. Anyhow, we can put any numbers we want into ... 5 Moving mass does generate gravitation different from stationary mass. This is the ''gravitomagnetic'' effect predicted by Lens and Thirring in the 20's and measured by Gravity Probe B: http://en.wikipedia.org/wiki/Gravitoelectromagnetism It is related to the ''frame dragging'' effect that you hear about with respect to spinning black holes. There, there ... 5 As the two charged bodies attract, they have unlike charges. So, Assuming your two charged bodies as conductors and charged equally, the system may be considered as a Capacitor. If you place a dielectric like glass of some Relative permittivity \epsilon_r (3.7 to 10) which fills the empty space between the bodies, then the capacitance would be ... 5 A charged particle circulating in a magnetic field does radiate energy, and it is called synchrotron radiation. All circular particle accelerators have energy losses due to this radiation. 5 The nature (and glory) of the dirac delta function is that the volume integral$$ \int_{\Delta V} dV' \delta ( \boldsymbol{r-r'} ) = \left\{ \begin{array}{cc} 1 & \text{if } \Delta V \text{ contains } \boldsymbol{r}\\ 0 & \text{if } \Delta V \text{ does not contain } \boldsymbol{r} \end{array} \right. $$Using this function, you can write the ... 5 If it is in air (or any other substance), there is a limit where the electric field of the object is going to be enough to ionize the surrounding medium, and the resulting current will drain the object of its charge. Similarly, if the object is immersed in vacuum, you will eventually have an electric field sufficient to "polarize the vacuum" by creating ... 5 The cross-section for$$ e^+ + e^- \to q + \bar{q} $$goes by the square of the quark charge (times the number of colors). Now, the quarks can not be observed in isolation because they hadronize. However the cross-section for$$ e^+ + e^- \to \mu^+ + \mu^- $$is identical except for going by the muon charge squared. So, a measurement of$$ R = ...

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Your intuition for the first one is correct; the extra charges are held to the surface by the electrostatic force, which is many orders of magnitude stronger than the gravitational force. The second one (and part of the first one): You're confusing free electrons with "extra" electrons. In a conductor, the highest energy electrons are not bound to any ...

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Once again, I am way out of my league in answering this. I may be wrong about many things here, comments appreciated That was just a definition of mass. The Higgs explains where rest mass (but not gravity) comes from in a mathematically rigorous manner. One of the attempts to explain how our universe works in a mathematically rigorous manner is the ...

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Because a proton can decay to a positron. It is an experimental fact that the proton and positron charges are very close. To conclude that they are exactly equal requires an argument. If a proton could theoretically decay to a positron and neutral stuff, this is enough. In QED, charge quantization is equivalent to the statement that the gauge group is ...

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On the level of QED and above, the equality of the charges has no theoretical explanation. But it is extremely well established experimentally, as even small deviations would add up to huge amounts of electricity in bulk matter. On the level of the standard model, the value of the charges of the up and down quark comes from simple arithmetic from those of ...

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Yes, however you will change the capacitance of the capacitor based on the dielectric properties of the glass you are using and the material you are replacing (which might be air or even vacuum). The general force between the two plates can be calculated as: $$F = \dfrac{1}{2}\dfrac{Q^2}{Cd}$$ Where Q is the charge, C is the capacitance and d is the ...

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If you followed the arguments carefully and checked what is demonstrably right and what is not, you would agree that what the argument actually does is to prove that a uniform electric charge density cannot have a uniform electric field. Your original task was to solve Maxwell's equations (well, Gauss's law), so if you find out that the equations aren't ...

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When physicists say that a particle has electric charge, they mean that it is either a source or sink for electric fields, and that such a particle experiences a force when an electric field is applied to them. In a sense, a single pair of charged particles are a battery, if you arrange them correctly and can figure out how to get them to do useful work for ...

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Physically what is happening is this: When you touch the positively charged source to the conductor (the metal sphere), electrons leave the conductor through the point of contact. This leaves the point of contact on the conductor with a large deficit of electrons, and thus the point has a positive charge density. The positive charge density produces ...

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In $\beta$ decay a neutron turns into a proton, an electron and an electro antineutrino. So if the proton and electron charge were not the same either the neutron must originally carried a net charge or the antineutrino must carry a charge. For the neutrino current limits are reported by the particle data group as less than 10$^{-15}$ of the electron ...

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The electric field of a negative point charge points towards the point charge as a result of the definition of the electric field of a point charge. To see this, recall that the electric field of a point charge $q$ is defined as $$\mathbf E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\mathbf e_r$$ where, $r$ is the distance to the charge, and $\mathbf e_r$ ...

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This is because the neutrality of polarity can be changed by electric field in this case. When you create - charge in the comb and you expose the pieces of paper to the electric field created by the charge, you will polarise them so that the part closer to the comb will be + and the other will be -. Here, see the electric field. The same polarities do not ...

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