# Tag Info

0

Q1. But how this can be explained "theoretically"? I assume your question is about the concept of 'electric potential ' due to a distribution of charges and in the present case 'a dipole'. The best way is to imagine an unit positive charge being carried/moved from infinity to a point on the equatorial line of the dipole. Naturally your probe charge ...

0

If you look at the electric filed lines you will note that a positive charge moving in the direction of the blue arrow does have a force on it but that force is always at right angles to its motion. Hence no work is done moving the charge.

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 ...

0

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 ...

0

To get A normalise the wavefunction. To get probabilities note you can expand any wavefunction in terms of energy eigenfunctions. Work out the co-efficients for each eigenfunction in the expansion by multiplying each side of the expansion with eigenfunction with quantum number n then noting that eigenfunctions with different n are orthogonal. the ...

0

So, the potential is not necessarily zero both at the ground and at infinity. First off, there's nothing in physics which forces the potential to be zero at infinity; it just happens to be a nice way to think of the Coulomb potential. Potential energies are not absolute numbers; there is a constant of integration that enters into them! Second, there is ...

0

The Gravitaional Potential Energy is calculated taking the potential at Infinty as zero, ie , if you come from infinity to the point at which you are calculating gravitational potential (say point P), Energy will continuously decrease because $$E \varpropto -\frac{1}{r}$$ (r is distance from the center of object to P) as r decreases, the fraction increases, ...

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 ...

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 ...

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). ...

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 ...

0

Even though it is called membrane potential, it's actually the potential difference between inner and outer cell, so in other words potential difference across the membrane. Since there is a potential difference of around -70mV in a standard animal cell (interior being the negative), it's normal that a bit more of the negative ions will arrange near the ...

0

The capacitance of a capacitor depends on the dielectric constant (relative permittivity) between the plates. The maximum charge that can be stored on a capacitor depends on the maximum potential difference across the plates of the capacitor which in turn depends on the maximum electric field (breakdown potential gradient - dielectric strength) which the ...

0

When you have capacitors in parallel they are all connected to the terminals of the source of the potential directly. Hence, they attract same amount of charges from the source if they are of the same capacitance. But if they are in series, then the reverse is the case.

-1

The more simple answer, I suspect is: you're looking at the electric field from a point charge ($\sim r^{-2}$) and not the potential, which goes as $r^{-1}$. If you simply write up your formula with potentials instead of electric fields, the answer is immediate.

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.

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 $$\int_{x^1}^{x^2}\sqrt{2m(E-V(x))} ... 2 Well, if there were charge between the boundaries, you would be solving Poisson's equation rather than Laplace's equation. However, boundary conditions for the potential function are also crucial, because there could always be distant point charges that modify the field in the region of interest, without changing the distribution of charge on the boundary. 2 The assumption you have to make is that the zero of potential is infinity. The total work done by unit positive charge in bringing it from infinity to point P is the potential at point P.$$-\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 = ...

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 ...

0

Since the charged conductor is present it will create an electric field and corresponding potential but to measure it experimentally you will need another known charge at one position, say A and then move it or allow it to move ( that depends on whether the charged body and our test charge are of same sign or the opposite sign) to a position B and then ...

0

You have to have probe wires in order to bring any charge into the voltmeter so it can be measured. Electrons will flow in a wire because they are attracted or repelled by other nearby charges. For example, if the charge is positive and you place the probe near the charge, electrons in the probe wire will move toward the charge, causing the other end of the ...

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 ...

0

The 6V in 6V battery is a label which gives an indication of the sort of voltage which might be obtained from such a battery. For example if your ^ V battery was an lead-acid battery and it was fairly new and fully charged its voltage would be 6.3 volt and if older or partially discharges then it is more likely to be 6 V. A 1.5 V alkaline battery at the ...

0

EMF or Terminal voltage will be considered same if battery has no internal resistance. If battery has some internal resistance then terminal voltage will be different(less) from the EMF or potential difference from the battery. If a battery has internal resistance then what will be considered if the same statement is given.

0

In your questions all the cases are assumed to be ideal unless mentioned. Therefore the electromotive force and the terminal voltage are equal in that case as internal resistance of the battery is considered negligible(if not given) . If the battery has internal resistance then the emf remains constant,but the terminal voltage decrease by a value which is ...

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 ...

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 ...

3

Boundary conditions of electromagnetic waves are well looked into because of the nature between polarization and reflection. However in this particular case it is rather simple, it is the evaluation of a contour integral over the boundary. This is done by summing the length elements of the pill-box shape that is typically used in this analysis in the correct ...

1

The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ...

0

First off, what is meant by the potential of the conductors? The "absolute" potential is of no importance in a working capacitor. In fact, it's practically impossible to determine. The potential difference between the plates (or between a plate and some ground/earth plane) is the important factor in the capacitance relationship, $C=qV$. Now ...

Top 50 recent answers are included