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Q1: I am not sure what you mean, physically it means the maximization of the the distances, but you already knew that. Q2: it is equivalent, so it is still an open problem. Q3: the formula does not apply, you are correct, for n=3 the solution is that the electrons reside at the vertices of an equilateral triangle about a great circle. But reading the ...

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Static electromagnetic fields implies: $$\frac{\partial\mathbf E}{\partial t} = 0 \quad\mbox{ and }\quad \frac{\partial\mathbf B}{\partial t} = 0$$ This means for electrostatics: $$\nabla\cdot\mathbf E = \frac{\rho}{\epsilon_0}, \quad \nabla\times\mathbf E = 0 \quad$$ And for magnetostatics: $$\nabla\cdot\mathbf B = 0, \quad \nabla\times\mathbf B = ... 1 I guess different authors use different definitions. For me, it is that the E- and B-fields do not have time derivatives, hence curl free, conservative E-fields and B-fields that can depend only on steady currents. The condition that the divergence of \partial {\bf E}/\partial t = 0 is not the same thing. The E-field could be time variable and have this ... 0 Electrostatic and magnetostatic are specific cases of the general electromagnetism. Defining a special case does not require to know a law/model that rules the phenomena. I don't need maxwell equations to define electrostatics or magnetostatics. I only need them if I want to know that my choice of special case is clever or useless. For instance, I can ... 1 Electrons are able to accelerate more freely then protons within a conductor. Therefore when a leaf electroscope is negatively induced the charged will move outwards while in a positively charged electroscope the leafs will stay hanging. 1 The original system as described in inherently unstable (or at best metastable). According to Earnshaw's Theorem, "a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges" http://en.wikipedia.org/wiki/Earnshaw's_theorem So the introduction of the slightest ... 1 Your boundaries are at r=a and r=b. Notice that the potentials at these two surfaces are independent of \theta (they are spherically symmetric). Look at a list of the first few Legendre Polynomials P_{l}(\cos{\theta}). For what value of l does P_{l}(\cos{\theta}) not depend on \theta? Further, notice that V(a) = V(r=a,\theta) = V_{0}, and ... 0 This structure consists of a series of potential wells at an atomic level: A quasi-free electron can travel trough this structure, where it is attracted or weakly bound by the individual atoms. In this case, a potential well due to an Al atom (or a virtual AlAs atom) is deeper than for Ga (or a virtual GaAs atom). In a picturesque view, the free electron is ... 0 First point is that you would normally get more charge at the edges. Second point - if you increase the gap then the capacitance would be reduce and the quantity of charge that could be stored for a particular potential difference would be reduced. Remember electric field is charge / distance so decreasing the gap would increase the electric field You ... 1 "Vacuum" actually means the ground state of space. When there are real particles in space, "it" is in an excited state, and no longer the vacuum. But we usually think of a small region of empty space as approximating a vacuum, for reasons of locality, etc. If you have two real particles in space, and talk about their mutual interaction, then you necessarily ... 0 Let's do some calculus. Suppose you have two plates, almost parallel (off by an angle \alpha). The plates lie in the XY plane, from (0, 0) to (x_1, y_1). At x = 0, the plates are separated by a distance z_0, and at x = x_1, the plates are separated by a distance z_1. We'll now consider an infinitesimally small element of both plates. (Since ... -1 A degree of closeness is called density, and number of field lines passing thorough any closed surface is one method to find density... 0 When two insulating materials are rubbed together, some charge will be exchanged between the two. That is, one will become more positively charged and the other negatively charged. To understand which one (rod or cloth) will become positive and which negative, we need to determine the relative position of these materials on the triboelectric series. ... -1 i think if the rod is only charged then its charge will be transmitted to the cloth whether its negative or positive .if both of them are charged then their charges will attract if they are of opposite sign and repel if they are of the same sign . hope this answer is enough . 0 It helps to think about Maxwell's equations here. The first one of which is$$ \nabla \cdot \mathbf{E} = \rho / \epsilon_{0} $$where \rho is the charge density. Electrostatics is always like a chicken and egg problem: charges make fields, but fields move charges around, so it can lead to some confusion. It turns out that in the steady state ... 0 Divergence can be thought of as the flux of a vector field per unit volume. It is positive if there is a net flux out of a small volume and negative if there is a net flux inwards. When you say "its diagram" - of course there are different ways of plotting vector fields. Perhaps the most common way is using field lines. In which case it can be ... 0 I think that the electrostatic dipole is an instructive analogy: Its net charge is zero, as well. But because the positive and negative charge are not at the same position, there is still a resulting potential and even an attractive force to other dipoles. So the suggested answer would be that for a perfect screening of the attractive force from the ... 0 You cannot achieve a stable system putting together a different number of positive and negative charges in a classical context. There will always be some force unbalance and some of them will be expelled from the system. You need to go quantum. Then particles are forced on energy levels that keeps them confined, so you can a huge variety of (molecular) ... 0 well, you could build a faraday cage and go inside it to look at your instruments, no need to transfer information outside - or you could place a recording device inside. if it's a problem for you in your mind that there is a void inside the conductor, that problem would apply on smaller scale as well, your instrument for measuring would still take some ... 3 What keeps electrons on a negatively-charged conductor from leaving? It is a quantum mechanical phenomenon. Wherever there exists an electric field potential there exist energy levels , i.e. stable orbital locations which can be occupied by an electron. How does this happen? Even the simple Hydrogen atom has a negative ion state, an anion. This is ... 3 I think that electrons do escape. For example, Electrostatic Ion Thruster: Electrons are emitted from a separate cathode placed near the ion beam, called the neutralizer, towards the ion beam to ensure that equal amounts of positive and negative charge are ejected. Neutralizing is needed to prevent the spacecraft from gaining a net negative charge. I ... 1 Electrons do leave surfaces either due to electrical potentials pulling them as you describe or as a result of a combination of heat plus electrical potentials. THe electrons are held in because the Fermi level is lower than the vacuum level. Furthermore, the difference in potential between the Fermi level and the vacuum level is 'felt' by the electrons ... 2 Why do not you try this? The shell is always an equipotential object. When the charge is inside the shell, the potential of the shell is kq/R with R the radius of the shell, so the energy of the charge is kq^2/R. When the charge moves to infinity, the potential energy of it is zero. Therefore, the work is kq^2/R. 1 Dust sticking to things is a complex process but can be broken down into several stages and analyzed. First though lets define our dust. Dust Size The aerodynamics of dust are most easily approximated by pretending all of the particles are spheres with a density equal to water (1000 \frac{kg}{m^3}). Each particle is assigned an aerodynamic diameter that ... 4 I think there is some trick? The electric field is horizontal thus the electric potential varies in the horizontal direction only, not the vertical direction. I wouldn't call this a trick but it does appear that the question tests your conceptual grasp of the relationship between the electric field and electric potential. In particular, you should be ... 1 Think of it like this. Electric fields point in the directions that positive charges will tend to go, which is towards a decreasing potential. So, if Q is positive, it will have radial electric field lines that will point along decreasing regions of electrostatic potential; that is, point A will have a higher potential than will point B, and so the ... 0 Since your question states "for a region of space", I'd say no... The electric field of a big sphere and the field of a smaller sphere are the same outside the radius of the biggest sphere (if the total charge is the same). 1 r_{12} is the distance between the charges. r_{12} is not the distance that the charge has travelled. To calculate the work done to move the two charges together we would use \int_\infty^a F.ds which is like the formula you have in your question - your formula works if F is constant - here we need to use an integral as the force changes as r ... 0 d can be infinite only in math books. Your statement about Coulomb's Law is true. if F=0 then q_1q_2=0. As d gets arbitrarily large F will get arbitrarily small. You can find a d that makes F as small as any number you choose ... except zero. I'm interpreting your question as being about Coulomb's Law itself, not any practical application ... 2 What Jim is talking about in the comments is a limit: for d becoming increasingly large, F becomes increasingly small. For example, if q_1 = q_2 = 1\,\text{C} and the charges are a distance d = 1\,\text{m} apart, we find that the force is$$F = k \sim 10^{10}\,\text{N}.$$If the distance is made 1000 times bigger, the force is$$F = ...

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Consider that the electrostatic potential $\varphi$ isn't directly observable. The potential $\varphi+C$ where $C$ is a constant gives the same electric field, and so the same physics. Because changing $\varphi$ by a constant should give the same physics, you cannot conclude that the charge on the shell is $4\pi\epsilon_0 a_1 \varphi_1$. The charge is a ...

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Your whole derivation is correct. Even in the presence of two different dieletric materials, the $\mathbf{D}$ field will not be affected, but for the free charge density that you already dealt with. So the field will be $$\mathbf{D}=\begin{cases}\dfrac{1}{5}\beta r^3\hat{\mathbf{r}},&\text{if r<a},\\ \dfrac{1}{5}\beta ... 4 The integral you wrote integrates \rho over the whole space. This is impossible to calculate if \rho is not known in the whole space. For example, when the charge \rho is known only inside some finite region enclosed by a metallic shell, the shell is known to have constant potential \phi on its inner surface. This information is useless in ... 1 We start with the integral$$\oint_{|\vec{r}|=R}\mathrm{d}\Omega\frac{\vec{r}}{|\vec{r}-\vec{r}'|}.$$Since we are integrating over \vec{r}, we can without loss of generality, arrange for \vec{r}'to lie along the +Z axis, so that \vec{r}'=r'\hat{z}. Then the angle between \vec{r} and \vec{r}' is the standard angle (in spherical coordinates) ... 0 To solve the integral, I came across the multipole expansion$$\frac{1}{\left|\vec{r}-\vec{r}_{0}\right|}=\sum_{l=0}^{\infty}\frac{r_{<}^{l}}{r_{>}^{l+1}}P_{l}\left(\cos\sphericalangle\left(\vec{r},\vec{r}_{0}\right)\right),\;\; r_{<}=\min\left(r,r_{0}\right),\; r_{>}=\max\left(r,r_{0}\right)$$With \vec{r}_0=\vec{r}^{\prime} one has ... 0 Fluorescent tubes normally light when a electron, produced by thermionic emission at a filament, inelastically collides with a mercury atom and raises the latter to a metastable state, which decays and emits UV light. The fluorescent coating on the lamp then absorbs this UV and fluoresces. In very high electric fields, the filler gas in the tube can become ... 0 I found the solution to my problem : The equations can be written :$$ I = V \times C_{ij} \times i\omega $$Where I is the vector of known currents going threw the electrodes, V the potential of the electrodes and \omega the rotational frequency of the current. In this case the potential of the transmitting electrodes is known and the current ... 1 Can we just calculate by using 1/2 CV^2? Yes. If v_C(t) is the instantaneous voltage across the capacitor, the instantaneous stored energy is just$$U_C(t) = \frac{C}{2}v^2_C(t)

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A battery produces the effect you ask about. It removes electrons from the metal that represents the positive pole (cathode) and brings them to the metal that represents the negative pole (anode). For doing this operation, energy is invested. But, you'd better read how that is done in Wikipedia under the title "Battery (electricity)". ...

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For simplicity, I'm going to assume a 0.7V diode drop in the ON state. In your diagram, you can call the low node ground (0V), in that case, you have one of two situations: The diode is on (closed switch) In this state, current can flow through the resistor and therefore, through the diode. In this case, 0.7V is dropped across the diode and the remaining ...

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The first expression refers to definition of polarization in terms of density of physical electric dipoles. The second expression refers to (approximate) physical law, which says the polarization is proportional to total electric strength.

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In the first equation $\Delta L$ itself is typically proportional to $E$. That establishes the connection between the two formulas. However, $\Delta L$ might not be proportional to $E$. It might be permanent. The first equation allows for this case, but the second does not. If $\Delta L$ is permanent, the material has a permanent electric dipole ...

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In solids the outer electrons of the molecules composing them form bands. Electrons exist in what are called Fermi level bands, the energy levels they occupy are practically continuous and little energy is needed for an electron to be removed from the surface, particularly in metals. Usually the material is neutral. If an electron is missing from a band, ...

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The diode has a central barrier at the PN junction which allows charge to flow across it only when a sufficiently positive voltage is applied to the P part with respect to the N part so that the barrier potential is overcome.In this case the diode is said to be forward biassed and conducts current.At all times when the P part has a voltage less than the ...

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Gauss's law can only easily be used in cases with high degrees of symmetry and where you can define surfaces where the E-field is either parallel or perpendicular to the surface vector(s) and is constant or is zero over that surface. In the case of a doughnut, there is clearly a high degree of symmetry, either along the axis of the doughnut itself or along ...

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From the information you have put in your question I think that in the reverse bias case the 'diode is open' means that it behaves like an 'open circuit' or 'open switch' and no current passes through it and thus no current goes through the battery. The load resistor has current passing through it that goes to circuit/components not shown on the right hand ...

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A dielectric effectively behaves as if it was thicker than it is. If the dielectric constant is $K$ and the thickness of the dielectric is $t$, then for calculating the force it behaves as if the thickness was $t\sqrt{K}$. To see this let's take the example we know about where the dielectric fills the space between the charges: In (a) the thickness of ...

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TEM waves do exist in multi-conductor waveguides such as coaxial guides, in fact they exist in any homogeneous waveguide with more than one conductor. There are no propagating TEM, TE or TM modes if the cross section is inhomogeneous but at cutoff frequency the hybrid modes degenerate to the respective transversal modes.

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The equation for capacitance is Q=CV or V=1CQ. I don't understand what is the physical meaning of this "C": Does the charge in a system changes linearly with voltage under all circumstances? This first part is the statement of the behavior of an "ideal" capacitor. Because it is an idealization, it is easy to characterize as a linear component whose ...

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Worth noticing the difference between capacity and capacitance. You want the latter to have the meaning of the former - when instead it just describes how much the voltage increases when you add charge. It doesn't tell us when the capacitor is "full" - for that you need to know the rated voltage as well as the capacitance. When a capacitor is used in a ...

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