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You start your question with a fundamental statement from mechanics, and end with an example from a very complex system. Newtonian mechanics fundamentally involves only pair-wise forces at a microscopic level. When you build up a more complex system, like that of a catalyzed chemical reaction, you hide the very complicated microscopic phenomena by ...


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I'm not sure whether you're asking for examples of interactions where three particles interact simultaneously, or for systems where pairwise interactions are not adequate to describe the dynamics. So I'll offer you some of both! Three-particle systems You're made of the product of a three-particle interaction: the triple-$\alpha$ process, where three ...


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-1 Volt change in voltage so change in energy is -2 Joule.Just use formula $\int \vec{g}.\vec{dl}$ for each component


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There have been attempts to model the body and study what happens during discharges, like from a taser. Finite Element Models (FEM) for testing human body are pretty accurate. Here's an example of current density distribution due to a shock to the skin and how the current flows. Where the physician's report described the body's response to current flow: ...


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Here is a circuit representing the system. $R_{wire}$ is the resistance of the section of wire between the bird's legs. $R_{bird}$ is the resistance of the bird (which you can measure by sticking the two probes of the multimeter to the bird's two feet - if the cable is insulated, you will have to add the resistance of the insulation as well). When the ...


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The potential difference between two points on a wire carrying a current is given by Ohm's Law, $V = R\cdot I$. Since wires used for long-distance power transmission have, by design, a very low resistance per unit length, and the distance between the two extremities of your hands is very small (~10cm), even for large currents the potential difference is ...


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I suppose you could view your situation as a circuit with two parallel resistors; those are a part of the cable and the bird. However the resistance of the cable is many orders of magnitude smaller than the bird's (the cable is effectively a short-circuit), so there won't be any appreciable current through the bird.


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Maybe the confusion here is between potential difference, Volts, and the field strength, Volts/meter. If the man is just hanging in air, the field strength is relatively low and his primary danger comes from birds pecking his eyes out. But if he hangs close to the ground, he's still at the wire's potential, but now instead of maybe 50 meters, he's 10 cm ...


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The only three-particle interaction I know of is a chemical reaction which needs a catalyst. Which other specific examples of three- or n-particle interactions are there? The most famous example is probably $H_2 + I + I \rightarrow 2HI$ Max Bodenstein proposed in 1894 this termolecular mechanism. It is confirmed that this occurs. See this article ...


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The pair interaction is only a convenient approximation. More detailed approximations need multiparticle interactions. The most well-known one is the Axilrod-Teller 3-body potential which gives the first corrections to the $r^{-6}$ pair interaction describing for small distances the van der Waals repulsion of neutral atoms. Another common way to model ...


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Generally speaking, this is a property obtained from the linearity of the problem allowing for a superposition principle, which will breakdown for nonlinear problems. E.g., particle hydrodynamic interactions can be described as above in a Stokes flow, but cannot in a Navier-Stokes flow or viscoelastic flow.


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Your teacher is right. Although you missed a factor of $2$ in last line. The case you are saying is NOT equivalent to this one. How do you plan on filling dielectric $K_1$ on the lower half which is already filled with $K_2$? This makes no sense.


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The line integral would only work for finding a potential difference, which I presume is what you need. Since the line integral needs a path(to perform the integration), hence, it would need a general expression for electric field(in your case, finding the potential requires integrating from $\infty$ to $A$, so you'd need the expression for all points ...


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The conceptual problem here is that of EMF, $\mathcal{E}$ vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge ...


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The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ...


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What you're calling the "dipolar component of the potential" is not actually that object. For something to be a 'component' of the potential it also needs to be a scalar; in that sense, the sum of those three terms, $$\Delta \phi=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})$$ could indeed be called the 'dipolar component', and it will work in ...


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The electric field is given by $$ E=-\nabla V-\frac{\partial A}{\partial t}, $$ where $V$ is the electric potential and $A$ is the vector potential. So you can get the electric field from the potential. Or if you have measurements of a large enough set of points that are close together compared to the scales over which the field varies you might be able to ...


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If $E< V(x) $ everywhere, and if we assume that the kinetic energy operator $T=\frac{p^{\dagger}p}{2m}$ is a (semi)positive operator, then the TISE implies $$ 0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0, $$ which is impossible. Here $H=T+V$ is the Hamiltonian operator.


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This particle with have an unphysical wave function which blows up (as can be quite easily derived). Therefore, in quantum mechanics, we do not have any particles with $E<V_\text{min}$.


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OP considers an equations of motion of the form $$\tag{1}\dot{\bf x}~=~{\bf B}({\bf x}),$$ where the vector field ${\bf B}$ is of the form$^1$ $$\tag{2} {\bf B}~=~{\bf \nabla}\times {\bf A}.$$ In other words, ${\bf B}$ is divergence-free $$\tag{3} {\bf \nabla}\cdot {\bf B}~=~0.$$ Eq.(3) is locally eqivalent to eq. (2), cf. Poincare's Lemma. Let ...


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My initial equation was correct by I had neglected to include the modulus signs. Thanks to @jazzwhiz for pointing it out subtly.


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The multipole coefficients associated with a $1/|r|$ distribution $\rho$ depends on the choice of origin. For example, if you have a point charge and you choose the origin to be at that point charge, then it will have a pure monopole character. However, if you choose the origin to be elsewhere, it will have nonzero expansion coefficients other than the ...


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Introduction $\def\ph{\varphi}\def\vr{\vec{r}}\def\eps{\varepsilon}\def\rmr{{\rm r}}\def\pd{\partial}\def\l{\left}\def\r{\right}\def\ltag#1{\tag{#1}\label{#1}}\def\div{\operatorname{div}}\def\grad{\operatorname{grad}}\def\nR{{\mathbb{R}}}$ You do not need a combination of a surface charge and a volume charge to re-produce a point-charge field outside a ...


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if the cube is an insulator or a dielectric, how do I find the surface charge density and the volume charge density? Since you are working with an insulator, you can have $uniform$ surface and volume charge densities unlike in the case of a conductor( other than a conducting sphere ). The reason is, excess charge cannot redistribute in an insulator as ...


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Yes, the uniqueness theorem guarantees it. If you find a solution in which the potential on the surface of the cube looks just like the potential of a point particle, then the potential outside the cube must be identical to that of a point particle since there is only one unique solution satisfying this this boundary condition. (Assuming no external ...


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Suppose you have a charged cube as shown below, Say it holds a charge of $Q$ Coulomb. Now, for this cube to behave like a point charge, you need to go very $far$ away from it such that at a particular position, your cube appears like a point charge. In this case, the cube $behaves$ just like a point charge with a charge of $Q$ Coulomb. It is clear ...



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