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5

First, the gauge invariance means that the solutions $A_\mu(x^\alpha)$ are not unique. For every solution, the gauge transformations of it are solutions, too. That may be a problem because sometimes we want to have specific values of $A_\mu(x^\alpha)$ that answer a physical question. Second, we sometimes gauge fix because the equations simplify. For ...


5

A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor. When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance ...


4

The problem is that they have too many solutions if the gauge is not fixed. Imagine you have some initial values, and want to solve it on the computer. Then you have to solve the equations for the next time step given the values for the previous one. But you have to compute the values for, say, four variables but have only three equations. You somehow ...


4

I think the answer is clearer if you consider the equipotential as shown in the diagram below. Forgive their straight line nature as they were easier to draw that way. Given that $\vec E$ must be perpendicular to an equipotential surface then in your computation of potential difference $\displaystyle V_{AB}=-\int_B^A\vec{E\,}\cdot\mathrm d\vec{r\,}$ the ...


3

Classically, angular momentum is only conserved in a central potential by considering the torque (correct me if I am wrong). In quantum mechanics, it is also true, isn't it? In QM, an operator is conserved iff it commutes with $H$, because $$ i\dot {\mathcal O}=[H,\mathcal O] $$ Therefore, the angular momentum is conserved iff it commutes with $H$. As ...


3

Remembering that resistance = $\frac V I$ work the resistance at $I=1$ and $I=2$. For the resistance to be constant the current-voltage characteristic must be a straight line and go through the origin. There is another parameter which is useful in some instances and that is called the incremental resistance $\frac {\Delta V} {\Delta I}$ which is related ...


2

The lower limit should be negative: $$ \int_{\color{red}-\sqrt{2E/k}}^0 \sqrt{2m\left(E-\frac{1}{2}kx^2\right)} dx + \int _0^{E/mg} \sqrt{2m(E-mgx)} dx =\left(n-\frac{1}{2}\right) \pi \hbar $$ but the answer is yes: this is the correct expression. The general expression for the WKB approximation is $$ \int_{x^1}^{x^2}\sqrt{2m(E-V(x))} ...


2

Potential refers to a particular point - or set of points which are "equipotential". So you can talk about the potential of one of the capacitor plates (because each is an equipotential surface) but not the potential of the capacitor (because when charged the $2$ plates are at different potentials). When talking about a capacitor, potential usually means ...


2

Well, if there were charge between the boundaries, you would be solving Poisson's equation rather than Laplace's equation. However, boundary conditions for the potential function are also crucial, because there could always be distant point charges that modify the field in the region of interest, without changing the distribution of charge on the boundary.


2

The assumption you have to make is that the zero of potential is infinity. The total work done by unit positive charge in bringing it from infinity to point $P$ is the potential at point $P$. $$-\int_\infty^r \frac{1}{4\pi\epsilon_0}\left(\frac{+q}{r^2}\right) dr -\int_\infty^{2r} \frac{1}{4\pi\epsilon_0}\left(\frac{-2q}{r^2}\right)dr = ...


2

Looks like, in general, $\frac{\partial E_x}{\partial y}\neq\frac{\partial E_y}{\partial x}$, so $\overrightarrow{E}$ cannot be a gradient of any decent function. Therefore, I don't think the problem has an unambiguous answer.


1

The diagram above has a very important feature. It is the connection between the Earth and the outer conducting shell. Assume that the Earth is a conducting sphere and has some net positive charge on it. This will mean that the outer shell connected to it will also have some positive charge on it but the wire between the outer shell and the Earth means that ...


1

The Lennard-Jones potential arises from the mutual polarisation of the two molecules. It is present for all molecules however it is a weak interaction and easily overwhelmed by electromagnetic forces. For example two ions would interact via their electric fields much more strongly than via the LJ mechanism and we'd get a $1/r^2$ force ($1/r$ potential). ...


1

If the force goes with $q/r^2$ and the surface area with $r^2$ it follows that the force is proportional to the area, since $r^2$ cancels out.


1

"Perhaps a sagging potential is one of ratios rather than sizes." Your conclusion is correct, but the idea is not restricted to the sagging well. This is a neat exercise in scaling, and it all comes from the choice of parametrization of the potential for given length $L$. To see this better in the present case, just rewrite the hamiltonian in terms of ...



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