# Tag Info

1

So, I assumed a pair of electron and proton to behave as a dipole Classically, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localized electron, but about orbitals. This is because particles have an associated wavefunction $\psi(t)$, which tells you the probability of finding your particle at a particular ...

4

Yes, applying an electric field does create a pH gradient and in fact you can observe this simply by adding a suitable indicator to your system. For example see the section Demonstration of pH Gradient Formation in this article.

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The energy of a dipole in an electric field is just the sum of the energies of it's constituents, and is easily seen to be $U = -\mathbf{p} \cdot \mathbf{E}$. From this you can derive the dynamics of the dipole through the generalized forces $-\boldsymbol\nabla_\mathbf{p} U$ and $-\boldsymbol\nabla_\mathbf{x} U$. This assumes that the dipole's constituents ...

1

A plane charge would be an infinite 2-dimensional sheet with constant charge density. Already in a line charge you have neglected edge effects, because the $1/r$ dependence holds true only near the line provided you are far away from the end-points. Similarly, for a plane, the constant electric field holds true provided that you are much closer to the plane ...

1

In vacuum (or everywhere else, really), Coulomb's law takes the form $\boldsymbol\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}$, whereas in a polarizable material it is convenient to use $\boldsymbol\nabla \cdot \mathbf{D} = \rho_\mathrm{free}$. The $4\pi$ vs $\epsilon$ has more to do with units. As for the sign, can you give a reference?

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Since the arrangement is symmetrical when you interchange N and S, I would say that the maximum strength will come when D2 and D3 are equal. Also, the smaller D1 becomes, the more the flux lines will become compressed. Also, a sheet of metal where you indicate would have almost the same effect as making the magnets wider, without an increase in mmf. So I ...

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The Landau Pole is not a problem for QED because at scales much smaller than it (the Planck scale, which is smaller than the Landau pole by 260 orders of magnitude) the (negative) gravitational self-energy of the particle will more than cancel out its electromagnetic self-energy. So string theory is not necessary in this case, just gravity.

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Interesting compactification, although I admittedly have no idea whether there are any mathematical pitfalls that one must avoid when doing such a geometric manipulation. However, assuming the apparatus is reasonably well-separated from nearby objects and that the apparatus isn't floating at a significantly different voltage than the surroundings, the choice ...

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The energy you seem to refer to is the electric part of the Poynting energy expression for some volume $V$: $$E_{\text{Poynting}}(t) = \int_V \frac{1}{2}\epsilon_0 \left|\mathbf E(\mathbf x, t)\right|^2 + \frac{1}{2\mu_0}\left|\mathbf B(\mathbf x, t)\right|^2 \,d^3\mathbf x.$$ The vector $\mathbf E(\mathbf x, t)$ in this expression is the electric vector ...

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