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The problem here is that you've failed to specify a boundary condition. Consider an electrostatics problem where you're given a charge distribution $\rho(\mathbf{r})$ and asked to find the electric field $\mathbf{E}(\mathbf{r})$. The electric field is the solution to the set of differential equations $\nabla \times \mathbf{E} = 0$ and $\nabla \cdot ... 4 It does give "Coulomb's law" with$\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$\lvert \vec r \rvert = r$. 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 ... 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 ...

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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. ...

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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 ... 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 ... 3 If you really mean "points", see the answer to this question. Basically, the logic is as follows: If you try to inject a finite current at a "point" in a bulk, it will necessarily lead to a divergent current density$\vec{J}$in a neighborhood of that point, proportional to$r^{-2}$(where$r$is the distance from the injection point.) A divergent current ... 3 Since Michael has already pointed out that the problem as stated has no answer, I will answer a different question instead: if we have a resistive spherical shell with inner radius$a$, outer radius$b$, and bulk resistivity$\rho$, and the surfaces of this shell are coated with a conductive layer, what is the resistance between the inner and outer surface? ... 2 The voltage becomes the same as Earth, but this doesn't mean that the charge goes to "zero". By "zero" here, I mean that the positive charges (nuclei) are perfectly balanced by the negative charges (electrons). You can call the voltage of Earth 0 Volts, but this is a relative measure. Charge, in the usage here, is not a relative measure because it is a ... 2 The statement means that the net electric field at any given point inside the sphere adds up to zero due to all the varying contributions by the charges on the surface. They exactly cancel out, and hence for any point inside the sphere, the value of electric field is exactly zero. 2 The fact is not only that is infinite, but that it is a plane. Consider the homogeneous charged point (i.e., a point charge):$E \propto \frac{1}{r^2}.$In the homogeneous charged (also infinite) rod:$E \propto \frac{1}{r}$. In the picture below, if you consider the effect of the yellow area of the plate right under a probe at a distance$d$, the ... 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 ... 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. 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 ... 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 ... 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 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$r$: $$V(r) = ... 1 Clarification from another source: Source: Physics For Scientists And Engineers, Paul A. Tipler and Gene Mosca, Sixth Edition, W. H. Freeman and Company, New York, 2008, p. 971, Fig. 28-20. I maintain that the loop will act the same as the bar. In other words, if you cut a thin slit down the center of the bar and less than the length of the bar (you ... 1 All you need to charge a battery from a capacitor is to have more voltage charged on the capacitor than the voltage of the battery. The size will only affect how much time the capacitor will charge the battery. If you could charge the capacitor over and over and discharge it into the battery every time it was full it would eventually fully charge the ... 1 If you have some (static) charge distribution \rho(\mathbf{x}) the the electric field is given by:$$ \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 $$In your ... 1 I stumbled upon this similar question. This OP is more general and I found the way to extend the answer. Apparently the plates are coupled to the environment through a capacitance as well. By superposition we have Q_1 = Q_{12} + Q_{10}, Q_2 = -Q_{12} + Q_{20}, Q = Q_0 = -Q_{10} - Q_{20}, where the index i = 0, 1,2 corresponds to the ith plate ... 1 The answer to your question involves the fact that one does not usually know a priori the electric field \textbf{E} (or, for that matter, its direction) of a charge distribution \rho. Gauss's law, in integral form, relates the flux of the electric field through some closed surface S to the charge enclosed within the volume bounded by S. Precisely, ... 1 As your teacher says, it holds for every surface, but a look at the law itself, should clear out why some form of symmetrie is desirable:$$ \iint_S \vec{E} .\mathrm{d}\vec{A}=\iint_S E . \cos\theta . \mathrm{d}A = \frac{Q}{\epsilon_0} $$Here, E and \theta are position-dependent, so to calculate the integral, you need to take care of a position ... 1 As an example, let us suppose that you want to use the Gauss law to evaluate the electric field generated by a body charged in an uniform way. The gauss law tell you that the flux over an arbitrary closed surface around your body is proportional to the total charge:$$\int_{\partial V} \vec{E}\cdot d\vec{S}=\frac{Q}{\epsilon_0} $$but this is an ... 1 If the sphere is initially uncharged, then the electric flux through is surface is zero, by Gauss' law. If we add just one point charge q', then we will find a net flux through the surface, which is wrong. However, add a second charge q^{''}=-q', then the net flux is given by the enclosed charge q'+q^{''}=0, and all is well. 1 The potential at the surface of a charged sphere is given by:$$ V = k\frac{Q}{r}$$so the voltage$V$is proportional to the charge$Q\$. Since the potential of the larger sphere decreases that must mean the charge on the larger sphere decreases. The same argument tells us that the charge on the smaller sphere must increase. So charge flows from the ...

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This is a good question because you are about to learn something. As outside charge create electric fields going in, it then also contributes fields going out on the opposite side of the surface. So they cancel themselves out. Where as the charge inside the surface only has electric fields going out.

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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 ...

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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 ...

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