# Tag Info

1

In case of varying magnetic fields the electric field is considered as nonconservative. Thus, it is simply not possible to define a potential for induced electric field (See also this post with answers). Alternatively, one can think of the induced current as charges moving in the circle of radius $r$. Through every part $r \,d\phi$ flows the same amount of ...

2

The EMF created by a changing magnetic field is not considered to arise from a potential. This can easily be seen because when there is an emf, a charge can move around in a complete circle and dissipate energy the whole way around, but a potential cannot drive a charge around in a circle, because potentials are conservative. The two pieces of the electric ...

1

This is a very simple problem to tackle. The points A and B are connected by a conducting wire and nothing else. Therefore, there is no potential difference from A to B. The magnetic field is a red herring. We are not told otherwise, so we assume the wire is an ideal conductor. That means there is never a potential difference across it. Remember Ohm's Law ...

1

Here, there is a time varying magnetic field at work, and it's flux through the given loop changes. Thus there is a non-Electrostatic Field induced along the wire, proportional to the rate of change of the flux. However, it is a non-electrostatic field", for example the closed integral over a path for this field isn't zero. Also, the line integral of this ...

2

Perhaps the easiest way to see that there can't be a potential difference between $A$ & $B$ is a symmetry argument. You're tempted to say that $A$ is at a higher potential than $B$ so that current will flow from $A$ to $B$. But continuing along the loop, I find that current must also flow from $B$ to $A$, which would lead me to conclude that $B$ is at a ...

2

A double well with a high or wide barrier will have a smaller $\Delta E=E_2−E_1$ than one with a low or narrow barrier. (Less coupling.) I think we can understand this intuitively as follows but first it has to be said that rob is right: the energies $E_i$ are NOT inputs but the eigenvalues of the Schrödinger equation. Width, height and potential of the ...

2

We don't control the allowed energies $E_i$ independently of the potential: the energies must be the eigenvalues of the Hamiltonian. The "inputs" are the shape and height of the barrier between the two wells. You can kinda sorta think of the energy difference between the symmetric state (with energy $E_1$ in your diagram) and the antisymmetric state (with ...

0

Electrons are Fermions so they are forbidden from being in exactly the same state. Two identical quantum wells placed an infinite distance apart will be identical because the wave functions do not overlap. However at the quantum wells are moved closer together the wave functions begin to overlap and the exclusion principle forces the energy levels to split ...

2

how do you find potential in a place where we have no intuition of force and are not allow to find it. Well I think this might be your problem; I've certainly never heard it said that you are not allowed to find forces. The Euler-Lagrange equations are simply another tool to finding the dynamics, but that doesn't mean you have to start from scratch and ...

2

I concur with Sam Weir, typo. $E= \frac {b}{x^2}$ So $\frac {6kv/m}{(1 m)^2} =6\,\mathrm{kV}$ at $1\,\mathrm{m}$ and $\frac {6kv/m}{(2 m)^2} = 1.5\,\mathrm{kV}$ at $2\,\mathrm{m}$ I see $6\,\mathrm{kV}-1.5\,\mathrm{kV}=4.5\,\mathrm{kV}$ with $x = 1$ being higher potential, regardless closer to the positive charge is going to be higher potential.

1

Let me arrange some information briefly. Cohesion-tension theory : phenomena that pulls water molecules at leaves producing tension + cohesion along entire stream of water molecules. Cohesion of water molecules mainly arises from high-strength hydrogen bond and the tension that presses the stream is generated from various mechanisms. On the other hand ...

1

The xylem actually creates a long thin string of water from the roots to the leaves of a tree. This string remains continuous by two forces. One is the cohesion of water molecules and the other is adhesion of water with xylem walls. Now in leaves water is evaporated. That decreases the pressure of water there. Because the protoplasm of those cells on leaves ...

0

In a general Lagrangian formalism, $L$ doesn't equal $T - V$. Rather it is a function of some field (be it scalar field, vector field, or whatever other field that is useful...) $\phi$, derivatives of $\phi$ and spacetime (x,y,z,t). This function is chosen so that the equations of motion produce the correct physical phenomena. In general, canonical ...

1

In this case, the battery is said to be "floating". Its potential with respect to earth can be suprisingly high or low. Small buildups of static electricity on the battery can easily charge it to hundreds or thousands of volts with respect to earth. The voltage difference across the battery's terminals is still $1.5\,\text{V}$, but the voltage of the ...

-3

In electricity and magnetism, we use the scalar potential to derive the electric field and the vector potential to derive the magnetic field because ∇⋅B=0 and ∇×E=0. IMHO there's a deeper reason than the mathematical expressions: the field concerned is the electromagnetic field. See section 11.10 of Jackson's Classical Electrodynamics where he says ...

3

Yes, we can define a magnetic scalar potential in some problems, specifically if the current density vanishes in some places. Note that the condition is not $\nabla \cdot B = 0$ since this is always true. To define the magnetic scalar potential requires that there be a quantity whose curl is zero (curls of gradients are zero), which is to say $\nabla \times ... 0 QUESTION Current is constant throughout the circuit with a resistor hence we cannot say that the electron loses kinetic energy after passing through the load. SOLUTION Current throughout the circuit with a resistor is constant , no doubt about that. But to be fundamental, current in a circuit is set up by the electric field, not by electrons. For ... 0 To maintain a current you need to "push" the charge through any obstacles on the path. If there is resistance against the current, then the "push" must be large enough to overcome this. The potential difference is this "push". Of course, as soon as the resistor has been passed, then a large "push" is no longer needed to make the current keep moving. Now ... 2 1) Why the potential at the surface? This approach is probably used because part (a) of the problem gives you an explicit expression for$V_S$and so the expression for$V(r)$is self-contained without having to consider what happens inside the shell. Basically the solution makes use of the fact that$\$ V_S = \int_0^{r_2}{E dr} = \int_0^{r}{E dr} + ...

0

Potential depends upon the dielectric medium. In your case air acts as a medium. Capacitance is directly proportional to Permittivity of free space, multiplied by the Area of the plate, and inversely proportional to the distance between the plates. The capacitance decreases with increase in distance.

Top 50 recent answers are included