# Tag Info

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To be precise, this is not a potential, but the potential energy of two particles interacting. Let us take for example two charged particles with charges $q_1,q_2$, separated by vector $\mathbf{r}_{2} - \mathbf{r}_{1}$. The potential created by the first particle is $$\varphi_1(\mathbf{r}) = \frac{q_1}{|\mathbf{r} -\mathbf{r}_1|},$$ whereas the potential ...

0

The potential difference for the terminal, not for circuit. I hope that's what you are trying to mean. It means that all of the cell potential is dropped across the internal resistance.

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Since you're not telling us what you've already tried and where you're stuck, I can only start with a general hint: Gauss law. What closed surfaces can you draw in this picture, and how much charge is in each of them?

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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.

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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: $$\frac{R'}{\rho R}+\... -1 Physicists love their Taylor expansions and especially expanding around a harmonic (square) potential. As one typically doesn't want a potential which is unbounded, potentials with the highest polynomial term being an odd power are rare and the next interesting one after the harmonic potential is the quartic one. One nice application in 1D is that you can ... 0 Yes, electric potential is a state function. It doesn't matter what path you take between two points in an electric field the total difference in electric potential will always be the same. The work done to move through the electric field however is not a state function. 0 Why, when a stable bound state exists, the energies of the related stationary wavefunctions are negative? First, we need to understand that Energy is relative, both in Quantum and in Newtonian mechanics. This means that different reference frames will determine different values for the energy of a given object at a given point in space and time. In this ... 0 We can write down the differential of the thermodynamic potential as (I don't go into the derivation it will be my starting point):$$d\Phi=dU+PdV-TdS \leq 0$$The thermodynamic potential Gibbs Free Energy is defined as$$G=U+PV-TS$$. For constant temperature and pressure the differential change is$$dG=dU+PdV-TdS$$Which is the left side of ... 3 Electron degeneracy pressure is not caused by any electromagnetic interaction. It is an "ideal" effect that would be present in any high density gas of indistinguishable, non-interacting fermions. By constraining the fermions to have high density (with gravity in this case), you force electrons to occupy states well above zero momentum and kinetic energy, ... 1 Your thoughts are essentially right, with this restriction: you should not think that what is emitted is anything physical, that actually will act on a body at arrival: these are potentials, after all, not fields. The resulting formulae for the fields are known as the Liénard-Wiechert formulae, and they are not really transparent as they are usually given. ... 0 Maybe get rid of the potential: it just complicates the issue. Can we get eigenfunctions to H_0=-\frac{d^2}{dx^2} with negative energy? Clearly, something like$$ \psi_0(x)=e^{-|x|} $$will do the trick, as far as satisfying the Schrödinger equation on either side goes. But what about the angle at 0? One way to fix this is to introduce a high ... 5 A Gaussian surface is a mathematical two-dimensional surface with zero thickness. You don’t integrate the field inside it. You integrate the flux of the field across it. 0 This is not always correct. For sources extending to infinity, the potential is not 0 at infinity! This holds for localized sources only. For localized sources, we are free to set any value at infinity and hence can be taken equal to 0. However, for sources extending to infinity, we can't choose arbitrary values at infinity since the source is also "present" ... 1 By definition the potential energy is chosen to be zero at infinity. It can also be defined to be zero at the ground. Generally speaking the work W done for moving a body against a force \vec F(\vec r) from point A to B is given by the difference of potential energies$$ W = U(\vec r_A) - U(\vec r_B). $$That is to say the potential energy at point ... 0 According to me potential energy is based on the interaction between two objects. Take for example gravitational potential energy which is a result of interaction of the body with the gravitational field of earth. Since at infinity, interaction between two objects is negligible, therefore the potential energy is considered to be 0 there. 0 Because any function that drops off as 1/r asymptotically approaches 0. 0 It is neither, it is a simple problem of electricity, known as Ohm's law. Voltage is just another term for (electrical) potential difference. Electrostatics is the theory of charges which are not moving. Electrodynamics is the study of how changing or moving electric and magnetic fields behave. Neither applies to Ohm's law. 0$$U=U(r(x,y,z))\frac{\partial U}{\partial x}=\frac{dU}{dr}\frac{\partial r}{\partial x}$$\Rightarrow$$\vec{\nabla} (U)=\begin{bmatrix} \frac{\partial U}{\partial x} \\ \frac{\partial U}{\partial y} \\ \frac{\partial U}{\partial z} \\ \end{bmatrix}= \begin{bmatrix} \frac{dU}{dr}\frac{\partial r}{\partial x} \\ \frac{dU}{dr}\frac{\...

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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 ...

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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 $$\sum_{i=1}^N ( {\bf F}_i^{(a)} - \dot{\bf ... 3 It is an application of the total derivative. If partial derivatives exist, the total derivative of a scalar function f of many (i.e. more than one) variables along a curve, x:t\rightarrow x(t), is$$ \frac{df}{dt} = \sum_i \frac{\partial f}{\partial x^i}\frac{dx^i}{dt} $$It is easily seen that this is true, because the change in f for a small change ... 4 The LHS expands to \nabla U(r) = \left [ \frac{\partial U(r)}{\partial x}, \frac{\partial U(r)}{\partial y}, \frac{\partial U(r)}{\partial z} \right ]^T The RHS expands to \frac{dU(r)}{dr} \nabla{r} = \frac{dU(r)}{dr} \left [\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z} \right ]^T Assuming, by some ... 1 Let me make some complements. What James Johns said about Helium atom is a quite a good example. Basically, if we are capable of solving the many body Schrodinger equation directly, then no non-local potential appears. We have no non-local interaction in the nature. However, if we want some effective models or make some approximations, sometimes we have ... 0 It sounds like you're concerned about the difference between the charge density within a non-zero volume (i.e. Q/V for some V \neq 0) and the charge density at a single point (with zero volume). But you can view one as the limit of the other. For example, if we consider some point \vec{r} in space, and we define Q(R) to be the amount of charge ... 0 First The integral form of Gauss's Law given by, For any closed surface$$\oint_S \mathbf{E}\cdot d\mathbf{a}=\frac{Q_{enc}}{\epsilon_0}$$Here it doesn't matter where the charge is located in closed volume and thus distribution can be anything ,Given a charge you can find the Flux. While in differential form$$\nabla\cdot \mathbf{E}=\frac{\rho}{\epsilon_0}...

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Without all the context I assume that the book is trying to make a distinction between a "real" battery, and an "ideal" battery. An ideal battery would deliver the same voltage regardless of the current draw. A real battery essentially has an internal resistance. Thus as the current increases the voltage decreases.

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You have an extra time derivative in your expression for $Q$, there should be just one. (Your present expression for $Q$ vanishes, for the same reason you have described in the question). Having made that alteration, if we demand that $\frac{dQ}{dt}=0$, we will need to substitute the equation of motion. Upon substitution, this turns out to be precisely the ...

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the initiation of breakdown depends on the availability of ultraviolet photons. by placing the electrodes ever closer together, it gets hard to get any UV photons into the gap to trigger the discharge, and the breakdown voltage rises.

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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 $0=\phi (r=a,\theta) =\frac{1}{4\pi \epsilon_0}[\frac{Q_i}{|a-d_i|}+\frac{Q}{|a-d|}]\\ \Rightarrow Q^2|a-d_i|^2=Q_i^2|a-d|^2\\ \Leftrightarrow Q^2(a^2+d_i^2-2ad_i\cos(\theta))=Q_i^2(a^2+d^2-2ad\cos(\theta))\\ \... 1 It has nothing to do with$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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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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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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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 ...

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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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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.

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