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2

You have misrepresented the citation in the book. The 5th edition page 757 discusses experiments with a hollow sphere and a solid sphere. The experiments verify that the exponent is 2 within experimental error.

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This mostly comes under the theories of matter and therefore essentially chemistry i.e the types of bonds. The typical bonds are metallic bonds, ionic bonds, covalent bonds, hydrogen bonds, as well as Van de Wall forces and the like. I advise looking into these. Electromagnetism is responsible for the amalgamation of elements into compounds of any phase, ...

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To my mind, the above explanation (and others commonly presented) is missing an important piece though. In the semi-classical intuition presented, there should never be a preference for spins to align. The reason is that Pauli exclusion slapped on top of a classical picture simply restricts the phase-space of the system, thus reducing entropy. Sure, the ...

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Much of chapter I, section 9 in "Foundations of potential theory, Oliver Dimon Kellogg, Berlin: Verlag von Julius Springer, 1929", parts of which appear in the answers by Mathaholic, Procyon and Qmechanic, is devoted to answer this question. Let $v$ be a small region of arbitrary shape, containing $P$ (defined by $\vec{r}$) in its interior. We consider the ...

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We put (four pi times) the permittivity $4\pi\varepsilon=1$ equal to one from now on (in the SI unit system) for simplicity. Let ${\bf r}_0\in \mathbb{R}^3$ be a fixed position. The electric field in the $i$'th Cartesian direction at ${\bf r}_0$ is $$\tag{1} E^i({\bf r}_0)~=~- \int_{[0,\infty[\times [0,\pi]\times[0,2\pi]} ... 1 I agree that the volume form in spherical coordinates cancels the 1/r^2 divergence from Coulomb's law, but I think there's a more physical way to interpret this mathematical fact: remember that strictly speaking, for a continuous charge distribution with no delta functions in the density, \rho({\bf x}) is not the amount of charge at point {\bf x}. ... 2 The electric field at a point outside the volume charge distribution is well defined, certainly. What about the electric field at a point inside the charge distribution? Let \vec{r'} denote a point that lies in the charge distribution and \vec{r} denote a point where the electric field is to be determined. The problem is the determination of the electric ... 0 I agree with gatso, but I think it might be explained more simply by noting that the volume element can be written as dV=4πr^2.dr which cancels out the r^2 in the denominator. I also agree that the problem is similar to integrating (1/√x)dx from x=0, because here the ratio dx/√x = 2√x.d(√x)/√x = 2d(√x). 5 I actually agree with Ruslan's comment. You cannot say that the integral blows up when \textbf{r} = \textbf{r}', with \textbf{r}' spanning the integration domain where \rho \neq 0. The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function. To see this you ... 3 I would say yes, the electric field is well defined in a volume charge distribution. You raise some interesting points, that I think ultimately comes down to: Is the electric field consistent and well defined mathematically? What is the physical interpretation of the mathematics? Is a volume charge distribution physical and realistic? To answer the first ... 4 The existing answers tell you why F= \frac{1}{2} QE is right, but I think it's important to say why F = \frac{1}{4\pi\epsilon_0} \frac{Q^2}{d^2} is wrong. Coulomb's law is not easily applicable here because the plates are not point charges. In particular, their sizes are not negligible (indeed, much larger) than the distance between them. It would ... 5 The energy of the capacitor is U= \frac{\epsilon_0}{2} S\,\mathrm d E^2 where S is the area of a plate. If we increase of \Delta d the distance of, say, the right plate from the left one, keeping fixed the charge Q on each plate, E does not change and we find a variation of energy$$\Delta U = \frac{\epsilon_0}{2} S E^2 \Delta d = ...

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The E field due to each plate is $E/2$ and hence the total field between the plate is $E$. But a plate won't exert force on itself, so the E field experienced by a plate is $E/2$ only. Multiplying by charge gives you the force, hence $1/2 QE$.

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You would expect that the force would not depend on whether or not the capacitor was connected to a constant voltage source $V$. If the capacitor $C = \dfrac {\epsilon_o A}{d}$ is connected to a constant voltage source the the energy stored in the capacitor is $U = \frac 1 2 CV^2 = \dfrac 1 2 \dfrac{\epsilon_o A}{d}V^2$. The energy stored in the capacitor ...

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Yes, because electric field lines travel at the speed of light in vacuum.However its effect at infinite distance would be negligible, as if field did not reach there at all.

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The range of the Coulomb force is infinite (the force between two charges $Q_1, Q_2$ separated by a distance $r$ is given as $F = \frac{Q_1 \, Q_2}{4 \pi \epsilon_0 r^2}$), with the implication that the photon has zero (rest) mass. However if you were to suddenly create (say) a positive then the "news" about this would travel at the speed of light, so that ...

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When doing electrodynamics, you don't really consider a finite distance at which field lines extend - you consider the behavior of the field as it extends to infinity. In other words, when you have a potential that's dependent upon a distance r from the origin, you take the limit as $$\lim \phi (\textbf{r})\rightarrow\infty$$ and see what happens (e.g. it ...

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The electric field from any single line charge always points directly away from the line charge, is proportional to the linear charge density, and falls off like $\frac{1}{r}$ In your problem, at any point of interest, you will have an electric field vector that has a component on the y-z plane from the line charge on the x axis, and a component on the x-z ...

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