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The divergence theorem is stated (using vector calculus notation) $$\iiint_{V} \vec{\nabla} \cdot \vec{F}\ dV = \iint_{\partial V} \vec{F} \cdot \vec{n}\ dS$$ For some $C^1$ vector field $\vec{F}$. If we consider, say, the electric field of a point particle : $$\vec{E} = \alpha \frac{\vec{e_r}}{r^2}$$ for some constant $\alpha$, then $E$ isn't ...

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A Gaussian surface is a mathematical two-dimensional surface with zero thickness. You don’t integrate the field inside it. You integrate the flux of the field across it.

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The LHS expands to $\nabla U(r) = \left [ \frac{\partial U(r)}{\partial x}, \frac{\partial U(r)}{\partial y}, \frac{\partial U(r)}{\partial z} \right ]^T$ The RHS expands to $\frac{dU(r)}{dr} \nabla{r} = \frac{dU(r)}{dr} \left [\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z} \right ]^T$ Assuming, by some ...

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Electron degeneracy pressure is not caused by any electromagnetic interaction. It is an "ideal" effect that would be present in any high density gas of indistinguishable, non-interacting fermions. By constraining the fermions to have high density (with gravity in this case), you force electrons to occupy states well above zero momentum and kinetic energy, ...

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It is an application of the total derivative. If partial derivatives exist, the total derivative of a scalar function $f$ of many (i.e. more than one) variables along a curve, $x:t\rightarrow x(t)$, is $$\frac{df}{dt} = \sum_i \frac{\partial f}{\partial x^i}\frac{dx^i}{dt}$$ It is easily seen that this is true, because the change in $f$ for a small change ...

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I do not understand the question. Is the question: "Why do I get a short circuit if I connect the red wire?"? The answer is you don't get a short circuit, instead the current will not flow through the resistor, R2. This is because Kirchoff's loop rule cannot be satisfied if any current flows through that path. Think about it, the voltage drop across R1 must ...

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The system can be written as $\phi (r,\theta) =\frac{1}{4\pi \epsilon_0}[\frac{Q}{|r-d|}+\frac{Q_i}{|r-d_i|}]$ Now for $r=a$ we want $0=\phi (r=a,\theta) =\frac{1}{4\pi \epsilon_0}[\frac{Q_i}{|a-d_i|}+\frac{Q}{|a-d|}]\\ \Rightarrow Q^2|a-d_i|^2=Q_i^2|a-d|^2\\ \Leftrightarrow Q^2(a^2+d_i^2-2ad_i\cos(\theta))=Q_i^2(a^2+d^2-2ad\cos(\theta))\\ \... 2 Surface charges on a conductor won’t make the interior equipotential unless the charge density is continuous. If it is discrete, as in your example, then the best you can get is a sort of multipole approximation to an equipotential surface. 2 Proceed as follows. First accept that the notion of a point charge (with a finite amount of charge but no volume) is itself a mathematical abstraction which may or may not be useful or appropriate in dealing with any given piece of analysis. Next, treat a small charged region, say a sphere, with finite charge density and finite fields. Finally, let the ... 1 Because your cylinder has finite height, you cannot get very far with separation of variables. You can do it of course (see another answer) but there's no reason to believe your potential field is everywhere cylindrical, and as you suspect you're gonna have trouble with the boundary conditions, especially on the "caps" of your cylinder. 1 It's a linear homogeneous PDE, so separation of variables should work here. Using the Ansatz$u(\rho,z)=R(\rho)(Z(z)$, cylindrical coordinates for$\nabla^2u$and ignoring the obsolete term in$\phi$we get: $$\frac{u_{\rho}}{\rho}+u_{\rho \rho}+u_{zz}=0$$ So that: $$\frac{R'}{\rho R}+\frac{R''}{R}+\frac{Z''}{Z}=0$$ Separation: $$\frac{R'}{\rho R}+\... 1 Your thoughts are essentially right, with this restriction: you should not think that what is emitted is anything physical, that actually will act on a body at arrival: these are potentials, after all, not fields. The resulting formulae for the fields are known as the Liénard-Wiechert formulae, and they are not really transparent as they are usually given. ... 1 Let me make some complements. What James Johns said about Helium atom is a quite a good example. Basically, if we are capable of solving the many body Schrodinger equation directly, then no non-local potential appears. We have no non-local interaction in the nature. However, if we want some effective models or make some approximations, sometimes we have ... 1 By definition the potential energy is chosen to be zero at infinity. It can also be defined to be zero at the ground. Generally speaking the work W done for moving a body against a force \vec F(\vec r) from point A to B is given by the difference of potential energies$$ W = U(\vec r_A) - U(\vec r_B). $$That is to say the potential energy at point ... 1 The issue is this: This would mean Chloride is flowing from a Low potential to a High Potential(When the natural order of things is High to Low) Positive charges move from higher to lower potential. However, since chloride ions are negatively charged, they actually want to move from lower to higher potential. This is because they want to move to lower ... 1 A problem with OP's derivation (v3) is the assumption that constraint forces have potentials, which is typically not the case. This can presumably be fixed by working with generalized forces rather than generalized potentials. After the above improvements, we claim that OP's equations will essentially boil down to$$\sum_{i=1}^N ( {\bf F}_i^{(a)} - \dot{\bf ... 1 It has nothing to do with$r’$being zero. The only place$r’$is zero is at the origin. When$\rho$is spherically symmetric, it is independent of the angular coordinates, say the usual ones$\theta’$and$\phi’$. Then, if one takes the$z$-axis along$\mathbf r$so that the angle$\alpha$between$\mathbf r$and$\mathbf{r}’$is just$\theta’$, the$\...

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Because it is a potential difference, it doesn't matter. For example: "I'm 10cm taller than my dad" and "my dad is 10cm smaller than me" contain the same information, just expressed differently. Similarly, suppose I measure $V_{AB}$ and say, "$V_{AB}$ is 5V", and you say "actually it is -5V", we still agree on the actual quantity, but just disagree on how ...

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The physics of both equations is the same. Look at the second equality in each one. Both agree on $V_B-V_A$. It’s only ambiguous notations $V_{AB}$ and $\Delta V$ that are confusing.

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For simplicity let's say $k_1 +k_2 = k$ . Now the spring is initially in its natural length . After the block strikes the spring it would come into SHM. So let us find the mean position for the SHM. $$k x_{mean} =mg$$ So $x_{mean} = \frac{mg}{k}$ Now we will see the situation as a case of SHM.The velocity of the mass at $x_{mean}$ from the mean position ...

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