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Here's the easy way to do it. Do Gauss's law for a single uniformly charged $yz$-plane with surface charge $\sigma$. You'll find that the electric field is distance-independent except for its sign; $$\vec E = \frac{\sigma}{2\epsilon_0} ~ \operatorname{sgn}(x) ~ \hat x.$$Here $\hat x$ is the unit vector in the $x$-direction and $\operatorname{sgn}(x) = x / ... 2 Complex analysis is very useful in potential theory, the study of harmonic functions, which (by definition) satisfy Laplace's equation. One way to see this connection is to note that any harmonic function of two variables can be taken to be the real part of a complex analytic function, to which a conjugate harmonic function representing the imaginary part of ... 1 If you have some (static) charge distribution$\rho(\mathbf{x})$the the electric field is given by: $$\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}$$ or it might be easier to calculate the potential using Poisson's equation: $$\nabla^2 V = -\frac{\rho}{\varepsilon_0}$$ and then calculate the field using: $$\mathbf{E} = -\nabla V$$ In your ... 0 There are a couple mistakes in your approach.$\rho A$is not the enclosed charge. To get a charge from a charge density you need to multiply by volume not area, and since$\rho$varies with$x$you really need to integrate. So say you take a Gaussian surface where one face is outside the slab and the other face is at$x=x_0$(and the remaining faces don't ... 0 It is not affected at all. There is no net potential difference across the sandwich whether it is part of a circuit or not. 0 Within a solid the electrons reside in energy bands. If we imagine adding electrons to the solid the electrons fill up these bands until the last electron added will have the highest energy. This energy, the energy of the most energetic electron, is called the Fermi energy, though at temperatures greater than absolute zero there is some energy from thermal ... 0 I think what Griffiths meant was that "Since Laplace's equation requires, on the contrary, that the sec­ond derivative is zero, it seems reasonable that solutions should exhibit no extrema." But it does not necessarily mean that a function cannot have a null second derivative in an extreme, since the$x^4$function has a minimal value at the same point ... 0 Charges would appear distributed in such a way that$E =\frac{1}{2}\int\rho(r^\prime)\Phi(r)dv $is a minimal. But$\Phi$is a function of$\rho$as well$\Phi = \int\frac{\rho(r^\prime)}{|r-r^\prime|}dv^\prime$So you have to minimize$E$with respect to$\rho(r^\prime)$subjected to the constraint of not going out of the conductor to find what ... 1 All you need to charge a battery from a capacitor is to have more voltage charged on the capacitor than the voltage of the battery. The size will only affect how much time the capacitor will charge the battery. If you could charge the capacitor over and over and discharge it into the battery every time it was full it would eventually fully charge the ... 0 Both good answers. Note that the Earth has a self-capacitance by the same arguments. The Earth carries a net charge and really does have a capacitance. Remember that capacitance is defined in terms of the work that you must do to take a charge from one plate to the other. Two charged objects are involved. Even if the plates are nearly an infinite distance ... 1 Clarification from another source: Source: Physics For Scientists And Engineers, Paul A. Tipler and Gene Mosca, Sixth Edition, W. H. Freeman and Company, New York, 2008, p. 971, Fig. 28-20. I maintain that the loop will act the same as the bar. In other words, if you cut a thin slit down the center of the bar and less than the length of the bar (you ... 0 What I think is that there is no net emf induced in the first case and hence no current in the loop but there is emf induced in the second case but as the circuit is not complete, there is no current. Here is my reasoning: case 1 in the first case the flux which is (B. A) is not changing and hence emf I. E. d(flux) /dt is 0. One can ... 0 Treat the individual electrons in each conductor as if they were in a closed container, and there was otherwise a vacuum in that container. When moving through the magnetic field in the top picture (the ring), I would expect electrons to move to the bottom of the ring, leaving a net positive charge at the top. This will only occur until the electric force ... 1 The atom has some charge distribution$\rho(r)$. We don't don't know what form the function$\rho(r)$has, but we do know it depends only on$r$because an atom is spherically symmetric. When you have a spherical charge distribution the potential at a distance$r$is simply due to the total charge inside the distance$r$: $$V(r) = ... 0 Let's deal with circuit elements (resistors, inductors, capacitors, batteries, meters) which have two connection ports. Two components are connected in series when they are connected to each other only once and not connected to other components at that connection point, also called a \textbf{node} (IMPORTANT term). If a third component is in series with ... 1 Coulomb's law is valid at all energies we have probed experimentally. There is a subtlety here because the fine structure constant changes with energy so the electrostatic force gets stronger with increasing energy, but it is still an inverse square law force. However a proton is not a point particle, so the force between for example a proton and electron ... 1 Yes, however the predominant forces are much, much stronger than electromagnetism. That is, there is a "strong nuclear force" which will routinely hold protons side-by-side in a tiny space, and a "weak nuclear force" which will routinely turn a neutron into a proton plus an electron, and those two forces completely violate what you'd expect from ... 0 Each individual atom is presumably neutral as well. So, we can simplify the model of a neutral object to a neutral atom and then go from there. Each atom has a positive core and a negative cloud of electrons around it as I'm sure you are aware. The electrons on average have a uniform field emitted in all directions and so does the nucleus. Because the ... 1 Both can happen simultaneously. This happens in binary star systems all the time. Work done on object A as it moves from position 1 to position 2 is calculated by$$\int_{\vec{r}_{1A}}^{\vec{r}_{2A}} \vec{F}_{total\,on\,A}\cdot\,d\vec{r}$$. That changes the kinetic energy of A. It could be negative which means the kinetic energy decreases. Some might say ... 0 Fundamentally you just need to know which direction energy flows. If energy flows from object (or system) A to object (system) B, object A is doing work on object B which is the same as saying object B is being worked on by object A. 2 Indeed, the \vec{E} field in a parallel plate is independent of distance from the plate. This works because of the assumption d \ll length of plate (thus, we can ignore side effects of the plate). And as Bort pointed out, it is the Voltage V that scales linearly with respect to distance from the plate, while \vec{E} will remain constant. 0 Yes the diagram in the book is not to scale. It should look like this: Each transition increases by 1V. The outer edge would have a potential of 1V, the transition between green and yellow would have a potential of 2V. The final transition has a potential of 6V. 0 Should it? Lets say the 6V mark is at distance at 1, then 4V is at 1.5 and 2V is at 3 meaning that 4V should be in in the middle between the origin and the 2V mark, which could work as far as i can tell on my screen 0 I do some work with magnetic fields in tissue as well for wireless power applications, though we don't typically deal with fields that strong hopefully I can help. First of all human tissue is largely magnetically transparent at low frequencies. While modeling the electromagnetic properties of tissue is very difficult problem (this is why you largely see ... 4 In the sea of electrons picture, the electrons in the conductor are not at rest: they are jiggling about like gas particles, colliding and changing direction constantly. You can think of them as billiard balls at zero gravity, confined in the volume of the piece of conductor at hand. According to thermal physics, their average kinetic energy is related to ... 0 It is easy to make a trap that is stable in some direction(s). Just not one that is stable against deviations in all directions. Your circular motion also encircles a region with charge, so you are not confining the circling charge within a charge free region. 0 If the sphere has a volume charge, it implies the sphere is a non conductor, in that case E field exists within the sphere(there being no scope for electrons to move under the effect of these field lines as it is a "non-conductor"). When you construct a cavity within this volume.. By gauss law, you get the closed Integeral of E wrt area to be 0 within this ... 3 You need to watch what you mean by the ambiguous term "derive", which can mean either "was derived historically" (i.e. was motivated by or is a derivative of, in the non-mathematical sense) or "is derived logically/mathematically". Historically, I think you are correct that \boldsymbol{\nabla}\cdot ... 0 Well I dont know if we can prove it but there is a much more elegant way of formulating EM which may be helpful here. As you may know there are two potentials on EM: the scalar potential \phi and the vector potential \vec{A}, from which \vec{E}(t,x) and \vec{B}(t,x) are derived. From this two objects and following symmetry considerations you can ... 0 As your first equation is correct, there is no difference in using RI^2 or U^2/R to calculate the power P = U \cdot I. My best guess to their choise is that the example gives you the resistance and the current, so this is the simplest way to calculate the power since you don't have to calculate the voltage in a middle step. Furthermore, it is often ... 3 Maxwell derived his equations from 1) charge conservation law; 2) Coulomb's law; 3) Bio--Savart--Laplace law; 4) Faraday's law of induction. The equation \boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0} was indeed derived from Coulomb's law and in its differential form is written using Gauss--Ostrogradskiy theorem. ... 3 No we cannot prove it; Maxwell postulated that it would hold dynamically because it made the most sense for it to do so as he pondered the displacement current problem. As you likely know, Maxwell pondered the inconsistency between Ampère's law for magnetostatics and the charge continuity equation. Ampère's law for magnetostatics reads \nabla\times ... 0 The distance dependance of the potential of an electrical \ell-pole goes as:$$ V_\ell(r) \propto \frac{1}{r^{\ell+1}} $$So a monopole has \ell = 0 and V \propto r^{-1}, a dipole has \ell=1 and V \propto r^{-2}, and so on. This is a very different distance dependance to the strong force. 1 The electrons do not even enter the wire, because the redox reaction between the substances in each of the nodes never occurs. Once the wire is connected to each of the nodes, electricity will flow through as electrons will be more attracted to the node with the greater reduction potential. 1 The electric field shows the gradient (slope and direction of change) of the potential. If the electric field is high magnitude, the potential is changing quickly with a change in position. If its magnitude is small, the potential is changing slowly with a change in position. If the electric field is zero, the potential is either at a maximum or a minimum. ... 0 Potential is not same as electric field,electric field is zero doesn't mean potential is zero too. your calculation is right,total potential is double the potential of each charge. Edit:For the 2nd part of your question ,there is nothing wrong in potential surfaces criss crossing like that(but electric field lines shouldnt criss cross like that). 1 To answer your question in one word, "Yes" Now, onto the explanation:- According to Faraday's Law, you will get a current in a conductor when the amount of magnetic flux linked with the conductor changes. Note that it is immaterial whether the source of the magnetic field is a permanent magnet or an electromagnet. All that needs to happen for you to ... 1 Both the zeroes are "real" to answer the question. There is absolutely no problem with multiple points in a space having the same electric potential. Note that we have ASSUMED the potential to be zero at infinity and based upon that assumption we have found out the potential to be again zero at the equidistant point between two opposite charges. So there ... 3 According to Faraday's law of induction,$$\mathcal{E} = -N {{d\Phi} \over dt}$$you will need a change in the magnetic flux \Phi in order to get an EMF or an electric field. So if you just put your coil in the magnetic field of the permanent magnet, you will not measure a current. There will only be a current, if you move the coil around so that the flux ... 0 Charges of the same polarity repel each other, so they are trying to maximize the distance between them because this minimizes the coulomb energy E_C that is proportional to the inverse distance r. ~~~~E_C = \cfrac{1}{4 \pi \epsilon} \cfrac{e^2}{r} In all of your drawings the charges would gather similar to the distribution in your first picture. The ... 0 When there is no \vec B field, the force density \vec f =\rho\vec E +\vec J \times \vec B equals \epsilon_0\left( \vec \nabla \cdot \vec E \right)\vec E. And if there is also no \vec B field in a time interval then \vec \nabla \times \vec E=\vec 0. This last part is important, because the time rate of change of \vec B is essential. But when ... 0 In your expression for E_x, q_2 should be minus. Plus is for positive charge & minus is for negative charge. The direction of electric field from a point charge is taken to be radially away from that charge i.e. positive-ness. The direction of electric field is taken to be radially towards the charge i.e. negative-ness. cos(45) is coming because, ... 4 It does give "Coulomb's law" with \frac{1}{r^3}, it gives it in its proper vectorial form$$ \vec E \propto \frac{\vec r}{r^3}$$which, when taking the absolute values, yields the form you are probably more familiar with$$ E \propto \frac{1}{r^2}$$since$\lvert \vec r \rvert = r$. 0 The right way to derive the total energy is to put the particles one by one. In a space centered at$e_1$and having nothing else, the energy to bring$p_1$from far away to a distance$r_0$is$E_{e_1,p_1}$, this you know how to calculate. The energy to put$e_2$at its position will have two terms coming from$e_1$and$p_1\$. By superposition, this ...

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The most straightforward way to see that there is no net charge on the inside surface of the conductor in the first case is to draw an imaginary closed spherical surface inside the body of the conductor and note that (by one of the various equivalent definitions of conductor) the electric field at every point of that surface must be zero. Thus the LHS of ...

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QED is rarely concerned with precise forces or fields, although you may calculate e.g. the Coulomb force in a non-relativistic classical limit, see my answer here. QED is a quantum field theory. As such, the electromagnetic field does not possess a definite "value" at any point, it is to be thought of as an operator-valued distribution and what you may want ...

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The answer to this question completely depends upon how the charges are distributed.Saying that Gauss law holds here is correct but not in all cases. Suppose that the positive charges are to be randomly placed. Such a distribution would have slight non-uniformity, which would develop further with time because positive repels positive. So, a net flux will ...

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This is only in case of a charged conductor in which no currents are flowing through it. Now, consider a gaussian surface inside the conductor everywhere on which electric field is zero (since there is no current). From Gauss' s law, we get q(inside) = 0 since E=0. Thus, the sum of all the charges inside the Gaussian surface is zero. This surface can be ...

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Suppose in an arbitrary shaped conductor, there are some unbalanced positive charges. Draw a Gaussian surface within the conductor surrounding the charge. From Gauss Law, we would find there is a net flux which is not possible inside a conductor.

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If the "excess charges" are in a conductor then they are, by definition, free to move. If there are excess charges distributed throughout the conductor then they will be compelled to move by any electric field within the conductor. Gauss's law tells us that if there is any net free charge within the conductor then this also produces an electric field within ...

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