New answers tagged

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As I read your statements, I get the impression that the difference between capacity and capacitance is not clear to you. The capacity of a capacitor is defined by its "physical" construction (length, width, area, volume, material, etc. C = kA/d). However, capacitance is a measure of how difficult/easy it is for a capacitor to store charge (C = Q/V , ...


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Electrons have a drift velocity which is very small. But electrons pass the charge. They do not flow with a charge on it. It's like dominoes that fall. The energy wave propagates through the falling dominoes, but the dominoes don't translate much. Also it doesn't matter who is propagating the charge. Electrons and protons and charged ions- all can do that. ...


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According to the rules of qft there are virtual photons in the vacuüm. No, according to QFT the vacuum is static, in the sense that $P^\mu|\Omega\rangle=0$. Or put it another way, The vacuum at a time $t$ is exactly the same vacuum at a time $t+\Delta t$ for any $\Delta t$. This means that the picture of particles constantly appearing and disappearing ...


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The Earth's surface is negatively charged and the ionosphere is positively charged. This results in a downward electric field of strength of the order of 100V/m at the Earth's surface. This is the "fair weather" value and assumes that there are no conductors close by. Well you are a fairly good conductor of electricity and so when you stand on the ground ...


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As another answer pointed out, your formula is for electric field around an isolated point charge. It doesn't apply to the case of parallel plate capacitor. Normally we use Gauss's Law to find the electric field between the plates of the capacitor. We know that the field between the plates will be uniform from the differential form of Gauss's Law ...


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You reference the equation giving the electric field near a finite point charge. There is no finite point charge in a capacitor (unless we count a single electron, but I think you'll find a single electron won't produce a very large field measurement on a human scale...). The charge is distributed uniformly, and as you (you're a test charge) get closer to ...


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there are lots of questions and explanations, why the field of an infinite plain is homogeneous and does not depend on distance. That is the approximation involved: that the plates are big. So outside the plates the fields add to zero, and inside it's double. No, charge is not brought from one plate to the other. If you have an alternating current, it will ...


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With electric and magnetic fields a charge can be set in motion. For example, one can accelerate the charge in circles or make it go up and down. Such an accelerated charge will produce EM waves. A way to think about it is that when the charge is in a fixed position it produces a field as given by Coulomb's law originating from the position of the charge. ...


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In my understanding, the free charge is any charged particle that is not being restrained in the boundary, while the bound charge is in the boundary.It does not matter whether the material you currently discuss is a dielectric or a conductor.


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For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ...


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I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$ Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ...


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We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ...


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You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...


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A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ...


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Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ...


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why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge $Q$ while the other plate has charge $-Q$ so there is no net electric charge stored. (2) If you increase ...


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The Stokes law equation for the drag on the oil droplet is: $$ F_d = 6 \pi \eta r v $$ wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer ...


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Although the situation is quite simple, and equations of motion can be written without much difficulty, these equations cannot in general be solved in terms of simple functions, except in a few special cases. Numerical solution is necessary in most cases, and 'orbits' will not be stable. Motion along the perpendicular bisector will be an oscillation but not ...


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(Ah, a very old question which is not yet answered... and I don't find a duplicate answer, either, though the problem is very nice ... the comment is point you there, and that the user "centralcharge" has edited the question is amazingly fitting, too :) but it's not yet elaborated) Yes, you can use the method of images. Normally you are right, it requires ...


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As stated by Lemon, electric flux through a volume enclosed by a closed surface is zero when the volume contains no net charge. Electric flux through a closed surface $\rm S$ is $$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.$$ Now, according to Divergence Theorem, \begin{align}\int_{\mathrm S} \,\mathbf E\cdot \mathbf ...


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The equation you provided is actually given by $$W=\frac{\epsilon_0}{2}\int E^2\,\mathrm d\tau$$ which is the energy stored in an electric field. This energy is utilized by the charge to generate it's field of influence or it's electric field. It's dependent on the magnitude of charge and the distance of separation between the charge and the point of ...


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We can assume w.l.o.g. that the electric potential $$\left. \Phi(r,\theta,\varphi) \right|_{r<r_0}~=~0$$ vanishes in the interior. As already argued in Daniel Mahler's answer, a surface charge distribution $\rho$ with support at $r=r_0>0$ is far from unique. In fact, the reader may check that any electric potential of the form $$ ...


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When there are no external fields the charge must be distributed uniformly. If you have an external field and the sphere is made of conducting material then it will act as a Faraday cage and the charges will distribute themselves to cancel the field inside the sphere, leading to a nonuniform charge distribution on the surface with a 0 field inside the ...


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Yes, this is guaranteed by the uniqueness theorem for Poisson's Equation, and in fact is more general than spherically charged shells. However, as another answer indicates, that if there are other charges present elsewhere, the charge on the sphere will shift to cancel off the field interior of the conductor.


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Do Weyl fermions carry electric charge? That depends on whether Weyl fermions exist, and whether they are what people say they are. See this from the article mentioned by John Rennie: 'Whereas electrons and all the other known fermions have mass, in 1929, mathematician and physicist Hermann Weyl theorized that massless fermions that carry electric ...


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We should probably start by pointing out that no Weyl fermion has ever been observed. The recent observations are of quasiparticles that behave like Weyl fermions. Speaking rather loosely (and at the risk of upsetting the QFT experts hereabouts) a Dirac fermion can be viewed as a sum of two Weyl fermions, and the observations are of paired quasiparticles ...


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"The divergence of any electric field created by any surface charge distribution would be zero." No it's not. Consider a charged conducting sphere with uniform surface charge density and a Gaussian sphere of radius greater than the original one. The electric field is diverging through the surface of the Gaussian sphere. So the divergence cannot be zero. ...


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The reason is that all experiments known can be explained by having two types of electric charge. To distinguish between the two types of charge them it is necessary to introduce labels, conventionally the labels were taken be "positive" and "negative". Because of history, electrons are given the label "negative".


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There is no structure of electrons as far as we know. It's a point entity. So it cannot be seen as something that has further structure or said to be having "parts". It's a fundamental particle


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Defining precisely what are all the quantum numbers is a difficult question because it depends highly on the model under consideration, even for the standard model. In particular any U(1) symmetry leads to a quantum number, and similarly some U(1) subgroup of non-abelian groups that commute with all other interactions can also be associated to quantum ...


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What do you mean by positive electrons? There are no positive electrons. Once an electron is lost from an atom, it's an ion, not a positive electron. The ions cannot move because they are tightly packed. It's the electrons that are free to move. Now the flow of electron from positive terminal to negative terminal is related to conservation of energy. The ...


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This is one of those questions which very much relies on the fact that the plane is infinite. So any electric field line which starts off going towards the plane will be counted as part of the electric flux through the plane. Any electric field line which has a z component must reach the infinite z=26 cm plane.


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http://www.britannica.com/science/electromagnetic-unit-of-charge electric charge Electric charge ...× 10−19 coulomb. In the centimetre–gram–second system there are two units of electric charge: the electrostatic unit of charge, esu, or statcoulomb; and the electromagnetic unit of charge, emu, or abcoulomb. One coulomb of electric charge equals about ...


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What happens when you touch an object with a positively charged object? Ans: It gets positively charged. Now, you have connected a semiconductor to a positive end of battery. What do you expect? Ans: Yes, it gets positively charged. Will the terminal pull electrons out of the doped silicon, or equivalently, inject holes into it? Yes, it will. ...


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The potential energy is related to a system not to an individual charge. So when you bring one charge towards another charge the system of two charges gains potential energy. What this means is that any work which is done by either or both of the charges will result in the loss of potential energy of the system (both charges).


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In your first case you said you were bringing a charge B to the first charge lets say A. Here your reference is A thats why u say you are bringing it to A. Think of it the other way, put you reference on B then your perception would be you are bringing A to B and now A will have a Potential energy. Thats how it works both ways. And Finally i would like to ...



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