25

Actually, we do! It's just that it's not the same "kind" of potential - and the reason for this is that magnetic forces work differently than electric forces. Magnetic fields, if you know, do not directly exert a force on charged particles, simply for being charged. (They would exert such on hypothetical "magnetically charged particles", but we've never ...


15

If the charge comes from a point of $V_3=0$ and gains kinetic energy, then it is because the value of potential energy at the new point is lower. Then potential energy can be converted into kinetic energy. Don't worry about the actual potential energy values - only the differences in value matter. Think of having a box on the floor. You might say that there ...


12

The potential is a kind of primitive function of a vector field, primitive in the sense of being the reverse of a differentiation, ie., an integral with a variable upper limit. The derivatives of the potential in all directions represent the vector field; not all vector fields can be represented that way but some do, for example the electrostatic field. Some ...


10

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.


9

As usefully pointed out in the comments, Ohm's law just states that voltage $V$ and current $I$ have a relation. Therefore, Ohm's law per se means equivalence and not causation. This could be not immediate to deduce, there is plenty of Physics laws which imply both a relation and a causation, but this is not the case for neither Ohm's law, nor Newton's ...


9

This post does not really answer your question, but still it is worth reading what I would like to highlight. Point charge refers to electrons or protons. But in no way does the electron seem to have a size, although it is still debatable today. So, what mathematics and our reality assumes today is that the electron is a point charge. A point is ...


9

There is a "magnetic potential" that appears in more advanced books and is defined as ${\bf B}= \nabla \phi$, just as ${\bf E}= -\nabla V$. (please excuse my equations if you are not familar with the $\nabla$ symbol) It's useful in regions where there is no current or no time-changing electric field, but it is "multivalued." Because of Ampere's law, if you ...


8

Here is an argument. On one hand, in 1D the $n$th bound state has $n\!-\!1$ nodes. But a bound state in the delta function well is in a classically forbidden region (and hence exponentially decaying) for $x\in \mathbb{R}\backslash\{0\}$, so there cannot be any nodes. Hence $n\leq 1$. On the other hand, any attractive potential in 1D has a bound state, so $...


8

Foundations In a closed system, the four combined laws of thermodynamics are as follows: $$ dU = TdS - pdV $$ $$ dH = TdS + Vdp $$ $$ dA = -SdT - pdV $$ $$ dG = -SdT + Vdp $$ You can obtain these in one of two ways. You can start from the first law (the first equation above) and apply the definitions of enthalpy, Helmholz energy, and Gibbs energy. ...


8

An eV Electron volt is equal to a volt This cannot be, as the election volt is a unit of energy, whereas a volt is a unit of potential (energy per unit charge). However, they are closely related.... So that volt it is equal to, is it based on 1 volt rms or one volt peak to peak or one volt average, or some other calculation of a volt? The electron volt ...


7

If I place a positive charge in between the 2 charges, it has electric potential energy because it is feels a force towards the negative charge and repelled by the positive and gains kinetic energy. It's true that a positive test charge feels a force towards the negative charge since a positive charge 'rolls downhill' in the (electric) potential, and ...


7

Expanding, we have: $$Ae^{ikx}+Be^{-ikx}=(A+B)\sin(kx)+i(A-B)\cos(kx)$$ Setting $A'=A+B$ and $B'=i(A-B)$ we have that $$Ae^{ikx}+Be^{-ikx}=A'\sin(kx)+B'\cos(kx)$$ so the two solutions are equivalent. It really doesn't matter in the end.


6

The potentials and field integrals require that the integrand be integrable, not continuous. Any continuous function is (Riemann) integrable, but an integrable function does not need to be continuous. In fact, for integrals in one dimension, a function is Riemann integrable if and only if it is continuous almost everywhere [that is, everywhere except on a ...


6

Earnshaw's theorem indeed holds in any number of spatial dimensions. The problem here is your assumption about the electric field. In any number of dimensions, the electric field obeys Gauss's law. So in $d$ spatial dimensions, $$E(r) \propto \frac{1}{r^{d-1}}.$$ In particular, for $d = 1$, the electric field of a charge is constant. This makes sense, ...


6

When dimensions become very small, we are no longer in the realm of classical physics where potentials have mathematical singularities as in the classical 1/r Coulomb potential. Point particles belong to the realm of quantum mechanics and there the laws and computational rules are different. The electron is a charged point particle in the standard model of ...


5

In circuitry, there are materials that conduct electric current. While that is useful, it is NOT the only way that currents come about: an electron beam in a vacuum, or a charged belt motor-driven in a van de Graaff machine, do not have a material conductor, nor any associated 'R' constant. If a conductor with one end grounded has voltage source ...


5

The formula $$ P_n = |\langle\psi_n|\psi\rangle|^2 $$ assumes that the pre-measurement state $|\psi\rangle$ and the observable's eigenstates $|\psi_n\rangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is $$ P_n = \frac{|\langle\psi_n|\psi\rangle|^2}{ \langle\psi_n|\psi_n\rangle\,\langle\...


5

You're right, both are solutions to the Schrodinger equation for this potential. They are equivalent representations using Euler's Formula. The preference for complex exponentials vs sinusoids comes down to mathematical convenience. They are generally interchangeable. As a rule of thumb, if the wave function is 0 outside of a bounded region of space, you ...


4

I don't understand how the delta-function can be a "well" in the first place. It's a spike! One way to think of the "delta-function potential" is as the limit of the finite potential well. We can consider a potential of the form $$ V(x) = \begin{cases} - V_0 & |x| < L/2 \\ 0 & |x| > L/2 \end{cases} $$ (Note that this is an overall shift in ...


4

This means the magnetic fields must produce electric fields inside the coil. Yes, but this electric field, called induced electric field $\mathbf E_i$ , is only one component of total field in the conductor of the coil. It is not the total field. Total electric field in an ideal wire of the coil is zero, but this does not imply that voltage on the ...


4

I am going to contradict the answers by @Bio (whose answer has since been deleted) and @lineage and say that $\text d\mathbf l$ is actually $\text d r'\hat{r}$ The other answers are mathematically correct, but it goes against our physical intuition with how the limits of integration are set up, as it seems you were discussing in the comments of the answer. ...


4

The reason it is possible to define an electric potential is that $\mathbf{\nabla} \times \mathbf{E} = 0$. We define $\phi\left(\mathbf{r}\right) = - \int_{r_0}^{\mathbf{r}}\mathbf{E} \cdot d\mathcal{\mathbf{l}}$, with $r_0$ as some reference point. This leads to $\mathbf{E}=-\mathbf{\nabla}\phi$. So, for a point charge $q$, we get $$\phi\left(\mathbf{r}\...


4

The total energy of a body is the sum of kinetic energy and potential energy. Kinetic energy depends on speed and potential energy depends on position. The wiki text is a special case of the fact that if we move a body from A to B without altering the kinetic energy then its energy difference is only potential.


4

We are interested whether a given force $$ {\bf F}~=~{\bf F}({\bf r},{\bf v},{\bf a},t) \tag{1}$$ has a velocity-dependent potential $$U~=~U({\bf r},{\bf v},t),\tag{2}$$ which by definition means that $$ {\bf F}~\stackrel{?}{=}~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}. \tag{3} $$ If we define the potential ...


4

There seems to be a basic misunderstanding between the electric potential energy of a system of charges and the electric potential at a point due to a system of charges. To make things easier let's assume that the zero of electric potential is at infinity and that the electric potential energy of a system of charges is zero when all the charges are ...


4

In a few words, it's because the vector potential is a polar vector, so it can't depend on the right hand rule. But if you give it a $\varphi$ component you have to pick one direction or the other, and the right hand rule is the only thing that can give you a preferred direction. More explicitly, being a polar vector means that it changes sign under ...


4

This can also be seen directly from Maxwell's equation, $\nabla \times \textbf{E} = - \frac{\partial \textbf{B}}{\partial t}$. $\textbf{E}$ will be a conservative field if its curl vanishes, which requires $\frac{\partial \textbf{B}}{\partial t}= 0$, i.e. when the magnetic field is static.


4

A velocity-dependent generalized potential $U=U({\bf r},{\bf v},t)$ satisfies by definition $$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.\tag{1} $$ If there is no velocity-dependence but possible explicit time dependence, this simplifies to the well-known gradient form $$ {\bf F}~=~ - \frac{\partial U}{...


4

I am going to focus on the last sentence of your question, which sums it up: "why are some points more probable than others?" The answer is that the waves have to satisfy Schrodinger's equation, and that equation includes that higher kinetic energy goes with higher $d^2 \psi/dx^2$. Meanwhile the boundaries of the box exert their influence, which is that the ...


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