# Tag Info

## New answers tagged potential

0

"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 ...

2

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 ...

0

What is the difference between the potential difference and potential energy of an electron? If I understand your question right, these terms are describing the same thing - one is just in a "per charge" version. Electric potential energy $U_e$ is the potential energy associated with one spot in the circuit. Electric potential or just potential $V$ is ...

0

Although the situation is quite simple, and equations of motion can be written without much difficulty, these equations cannot in general be solved in terms of simple functions, except in a few special cases. Numerical solution is necessary in most cases, and 'orbits' will not be stable. Motion along the perpendicular bisector will be an oscillation but not ...

0

If it wouldn't be its tangential component would exert a force on the charges and they would move. The condition would then not be static. After some movement and redistribution of charges (when there would be no force on charges) the condition will again become static thus making the field only normal to the surface

0

I believe that the potential at point X should be 12−60I and the potential at Y should be 12−30I. However, in the several problems that I've tried, this is not the case. Please explain, why? A proper way to write this is (in terms of the node voltages and branch currents) is $$V_X = 12V - I_X \cdot 60 \Omega$$ $$V_Y = 12V - I_Y \cdot 30 \Omega$$ ...

0

If the wire has a non-zero resistivity, then there will be a finite resistance between the points $A$ and $B$, say $R_{AB}$. However, to determine the p.d. between these points you would have to know the resistance of the whole wire (or do you just mean that the wire between the points $A$ and $B$ has finite resistivity?), otherwise you can use Ohm's Law to ...

2

It is sometimes easier to visualise what is happening by using the idea of potential. To do this make one point in the circuit 0 V. This is a totally arbitrary choice. It is the bottom right hand corner of your circuit. Note to make the sums easier I have change the emf of the battery to 90 V so 2 A flows through the battery and 1 A through each of the ...

2

The electric potential of a point in space is defined by the mechanical energy that it takes to get a (positive) unit charge to that point. We usually define the potential of an infinitely distant point as zero, and then the movement of our test charge is from infinity to the point for which we want to measure the potential. In case of a capacitor the ...

3

Yes, it is perfectly possible to satisfy more than one gauge condition. The easiest way to see that is to consider a static configuration of fields. In a static problem, all quantities are time independent. That means that the Coulomb gauge $\nabla\cdot\vec{A}=0$ and Lorenz (not "Lorentz"; it's named after a different physicist) gauge ...

1

It's called the virial theorem: in a bound system, the average of the potential energy of a $\frac{1}{r}$-potential V is related to the average of the kinetic energy T like $\frac{\langle V \rangle}{2} = - \langle T\rangle$.

0

The tangential velocity is related to the radius of a circular orbit and the angular velocity of the orbiting object. Specifically: $$\overrightarrow{v}=\overrightarrow{\omega}\times\overrightarrow{r}\to v=\omega r$$ For the satellite, the centripetal acceleration is equal to $r\omega^2$, which is the same (according to the above relation) as the tangential ...

1

The relation is indeed correct. Let me show you why. This type of problem is solved by equating forces. Other classical mechanics problems are by equating energies or, in rare cases, momentum. The gravitational potential indeed is $$V(r) = G \frac Mr \,.$$ This is not directly useful for this problem. Here you rather want to look at the gravitational ...

0

First of all, electrostatic field is conservative in nature. So this force can be written as the gradient of a scalar potential V $E=-∇V$ The d in your equation is not right. The gradient indicates the rate of change of V with respect to the three coordinates along the three directions. Since the electric field is zero, we can write $∇V=0$ which ...

0

"Potential at a point is sort of energy you used to move a point particle."It is work done to move a unit positive charge in a field. Consider a shell with radius r and charge q. Potential at infinity is taken as 0 by convention. When you move a charge from infinity to surface of shell , you did work $W = \frac{kq}{r}$ in present of electric field E. ...

0

I) The Dirac delta distribution (and derivative thereof) in the dipole field $$\Phi ~=~\frac{1}{4\pi\varepsilon}\frac{\vec{p}\cdot \vec{r}}{r^3} \tag{1}$$ $$\Downarrow$$ $$\vec{E}~=~-\vec{\nabla}\Phi ~=~ \frac{1}{4\pi\varepsilon}\frac{3(\vec{p}\cdot \vec{r})\vec{r}-r^2\vec{p} }{r^5} -\frac{\vec{p}}{3\varepsilon}\delta^3(\vec{r}) \tag{2}$$ $$\Downarrow$$ ...

0

The credit to this answer goes to @knzhou, who provided it to me in the chat. What earthing does is simply equalizes the potential of the Earth (as in the planet, or the ideal earth, whichever is relevant). Now, we see that to calculate the potential difference between the plate (or whatever it is to which the earth is connected) and any point on or in the ...

0

There are two equivalent definitions for the potential difference between two points. The potential difference between two points is the work done by an external force in moving unit positive charge from one point to the other. The potential difference between two points is the minus work done by the electric field in moving unit positive charge from one ...

1

As Giorgio says, you are overcounting the terms when multiplying by 4. I would like to elaborate a little bit, just to show how this is more than just a math error. It is also a conceptual error: You mention in your reasoning "The total potential energy = the sum of the PE of each of the 4 charges." But this is false. Not only that, but there is no such ...

1

This time you're wrong. You are overcounting the terms when multiplying by 4. Potential energy is given for two particles at a time, so you should be multiplying by 2, which is the same as multiplying by 4 and then dividing by 2.

15

The fact is that, in the general case $$\vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t};$$ (signs depend on conventions used) where $\vec{A}$ is called vector potential. You can consult for example Wikipedia. Let us consider homogeneous Maxwell equations: $$\begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + ... 1 The mass of a ball is scalar. Its potential energy is stored in its position in a gravitation field. A dipole has its potential energy in its orientation with regards to an external field. It can do work by exerting torque when orienting along the field. 2 For dynamic electric and magnetic fields, there is a piece of the electric field that depends on the vector potential:$$ \vec{E} = - \vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B} = \vec{\nabla} \times \vec{A}. $$Taking the curl of the first equation yields Faraday's Law (with the V-dependent term dropping out as you note); taking ... 6 When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form \mathbf{E}=-\nabla V. 0 I want to find out the potential energy of a electric dipole in a uniform electric field by another process which gives the result to U= -P.E. Purpose A, B are the position of the point charges q and -q and seperated by small distance D and O be the position of the origin. The distance from O to -q is R. So potential energy, U=-q.v(R) q.v(R D) ... 0 So I actually integrated this a while ago, I thought I should share the result. It turned out rather simple with the right substitution. By the way I'm talking about both potential and charge disrtibution as a Debye-like. Point charge is straight forward... First we put the center of the charge density on the z-axis as mentioned in my question. I get$$ ...

-1

You can think of the arrangement as two parallel capacitors with a common connection $B$ which has a charge $+Q$ on it for all time. The capacitance of $BC$ is twice that of $AB$. Equal numbers of positive and negative charges are induced on plates $A$ and $C$. When both $A$ and $C$ are connected to earth they are then at the same potential and so the ...

0

What actually happens on earthing something in electrostatics? Grounding a charged rod means neutralizing that rod. If the rod contains excess positive charge, once grounded the electrons from the ground neutralize the positive charge on the rod. If the rod is having an excess of negative charge, the excess charge flows to the ground. So the ground behaves ...

0

To place a charge in the vicinity of an electric field, you should do work against the electrostatic force on the charge. This work done to bring a charge q to an electric field of some other charge configuration from infinity to a distance r, in the field is what we call the potential at the point r. To do a work to move a charge q from a potential V to a ...

1

Relation between Electric field and potential The relationship between electric field $\bf E$ and scalar potential $\varphi$ is given as $$\mathbf E= -\mathbf \nabla\,\varphi$$ where $\mathbf \nabla \equiv \textrm{gradient operator}\;.$ I am unable to understand from this - sign comes. It is worthy to quote from Purcell: The minus sign came in ...

0

As written with the bottom two being charged, there is no motion because they are still tied together. Presumably you want the left two to be charged so they can move away from each other. The CM does not move because every force in the system is balanced by an opposite force-there is no net force on the set of masses. This observation is useful because ...

0

First of all, we are fundamentally interested in E & B. But essentially, it comes down to the fact that only E and B are physically measurable and so $\phi$ and A are mainly only considered as mathematical constructs, but this isn't always true - they can be conceptualised. I apologise that I cannot give you an immediate physical insight, but I can ...

0

The vector potential has a divergence of zero; we can obtain some intuition by considering the geometry required by the divergence theorem: the volume integral of the divergence of the vector potential is zero for any volume, hence the total net flux through any surface is zero. So given your specific conditions, you can imagine geometric boundaries and ...

0

Electric field lines are always at right angles to equipotential lines or surfaces. The electric field is minus the potential gradient. So in the diagram showing a uniform electric field a positive charge would experience a downward force in the direction of decreasing electric potential. In this case the magnitude of the electric field is $\frac ... 0 The electrical field is related to a force concept: it describes the force per unit charge. The potential is related to a potential energy concept: it's the added electrical potential energy per unit charge. So, just as the force is the negative gradient (or in 1-dimension, the negative slope) of the potential energy function, the electric field is the ... 0 It looks like you are confusing electric potential with electric potential difference. The electric potential difference is a measure of how much the electric potential will decrease (or increase). If the electric potential decreases across a bulb that doesn't mean the potential difference decreases it means that there is a potential difference. Imagine ... 0 If the grey and blue parts of your diagram are the conductors and the magnitudes of the two sets of charges are the same then you have drawn a correct diagram with an electric field present only in the region between the two sets of charges. The field in that region being the same as if there was a$-Q\$ charge at the centre of the arrangement. In all other ...

0

If you are talking of a conducting sphere, this situation cannot happen physically. Charge is supposed to stay on the outer surface of the conducting shell. However if you make the shell infinitely thin (which is again not possible physically,) then the potential inside (for radii less than the radius of the shell) would be still a constant.

0

Grounding doesn't set the charge to zero - it sets the potential to zero. That means the residual charge will be whatever it needs to be to achieve that.

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