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The correct answer is "It depends, and the way it's shown, it's neither!". The electric field around a wire depends very much on the electrostatic boundary conditions of the problem (you are showing a ground connection... what, exactly, is that connected to?) and on the frequency of the AC source relative to the resonance frequency of the antenna. Unless you ...


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“[A]n equal and opposite field” is an oxymoron and gibberish. Henceforth I’ll assume that we have a chemical voltage source there. The Coulomb’s law is applicable in the case of a charged body, a body that has a constant electric charge. A chemical voltage source is a completely different situation: there is no pre-defined distribution of charges (as they ...


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How should it be “very weak”? Its field strength is immense, against macroscopic samples. Sure, one shouldn’t suppose a Maxwellian E-M fields inside a matter, especially an electric conductor – this microscopic field is uncertain. One hardly can understand it thinking about it as a vector or tensor field in the spacetime; one needs QFT. There is a related ...


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Electric field is a vector quantity. So, treat them as vectors and find the vector sum of the electric fields. $$\vec{E}_{net}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_{3}$$


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Hint: The divergence theorem tells us that the divergence of a vector field integrated over a region $R$ with boundary $\partial R$ equals the integral of that vector field dotted with the outward-pointing normal along the boundary; \begin{align} \int_R dV\, \nabla\cdot \mathbf v=\int_{\partial R} dS\, \mathbf v\cdot\mathbf n. \end{align} If we are ...


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From Gauss's law & using the conservative field, we have $$ \nabla^2u=-\frac\rho{\varepsilon_0} $$ In spherical coordinates and assuming isotropy in $\theta$ and $\phi$ directions, the above becomes, $$ \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)=-\frac{\rho}{\varepsilon_0}\tag{1} $$ Since we are dealing with a ...


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Wikipedia takes a swing at explaining this phenomenon. There it says that the velocity saturation is caused by scattering from optical phonons, with $$ \frac{1}{2}m*v_s^2 \approx \hbar \omega_o $$ where $m*$ is the effective mass of the carrier (depends on conduction band), $v_s$ is the drift velocity in saturation and $\omega_o$ is the angular frequency of ...


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I think I can clear up most of this. Maybe someone whose qm chops are better than mine could help with the parts I'm fuzzy on. Suppose an even-even nucleus has a prolate deformation (like an American football). This is very common, and basically occurs for any nucleus whose N and Z are both far from any magic numbers. What we really mean when we say that ...


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In a metallic conductor the most energetic conduction electrons are moving at the Fermi speed, which is surprisingly high. For example in copper it's around $10^6$ m/sec. However the electrons are continually scattering off the lattice of metal atoms so the direction of the electron motion is randomised and the average velocity relative to the lattice of ...


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There is a sort of analog called gravitomagnetism (or gravitoelectromagnetism), but it is not discussed that often because it applies only in a special case. It is an approximation of general relativity (i.e. the Einstein Field Equations) in the case where: The weak field limit applies. The correct reference frame is chosen (it's not entirely clear to me ...


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There is a gravitational analogue of the magnetic field. See gravitoelectromagnetism and frame dragging on Wikipedia.


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I solved my problem numerically, using the diffusion equation $\frac{\partial V}{\partial t} = -k\nabla^2 V$, with the following boundary conditions: Voltage at point D is fixed at 1.0 Voltage along the vertical line halfway between points A and D is fixed at 0.5 (voltage at point A is 0.0, use symmetry so we don't have to simulate the left half of the ...



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