# Tag Info

0

The way I remember these things is that gravitational potential energy is negative (i.e. in a situation where the particles attract each other the potential energy is negative). If the particles repel (two positive charges or two negative charges) then the potential energy is positive.

-1

We know that field due to point charge is zero at a distance infinity (and beyond) from it. Let us consider a square with equal charges at the corners. Let us assume that the length of the side of the square to be such that, electric field due to charges at corners of that side, becomes zero (as like field is zero at infinity and beyond it) at some ...

9

In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where $\mathbf{E}=\mathbf{0}$. Letting the charges be located at $(\cos(2\pi k/N),\sin(2\pi k/N))$ for $k\in\{1,2,...,N\}$, we can generate plots of $|\mathbf{E}|^{-1}$ for various $N$. The zeros of ...

8

One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...

1

This is because it is assumed that the test charge does not produce any electric field of its own and its magnitude is negligibly small, so it doesn't apply any force on the test charge.

0

http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-7-capacitance-and-field-energy/ Watch this video from 24:00 minutes till about 32:00 minutes (a total of about 8 minutes).

0

The magnetic field polarizes orthogonal to the electric field in free space. We generally only talk about the electric field because Maxwell's equations define a one to one relationship between the two. It would make just as much sense to only talk about the magnetic field. We choose the electric field because, in general, when light interacts with matter ...

0

The magnetic field does not vanish when light is polarized. A changing electric field induces a magnetic field, and a changing magnetic field induces an electric field. This is why, in the propagation of an electromagnetic wave, there is always an oscillating electric field coupled with a magnetic field oscillating perpendicular to this electric field. You ...

1

It does make sense to talk about the number of field lines, but only if you take care to represent the field amplitude as being inversely proportional to the spacing between the lines. With that, the total number of lines crossing a surface is proportional to the flux, etc. Some people, notably textbook authors Chabay and Sherwood, feel that the field line ...

0

Before answering your question i will state two maxwell laws These laws are intricately linked. A changing magnetic field instantaneously produces a changing electric field but with displacement current (J). Once electric field and displacement current are produced the cycle ends owing to conservation of energy. The same happens with changing electric ...

-1

Your problem is clearly and comprehensively treated by Hans De Vries in: http://chip-architect.com/physics/Magnetism_from_ElectroStatics_and_SR.pdf The quintessence is that a current carrying wire appears electrostatically charged to an observer in relative motion to that wire, even when the same current carrying wire appears uncharged to an observer at ...

2

What you're calling the "dipolar component of the potential" is not actually that object. For something to be a 'component' of the potential it also needs to be a scalar; in that sense, the sum of those three terms, $$\Delta \phi=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})$$ could indeed be called the 'dipolar component', and it will work in ...

1

from a, b and c: An electromagnetic field is propagating by changing the field that's generated by electrically charged particles pass through the air and the space that is devoid of particles of space. An electromagnetic wave propagates, not an electromagnetic field. An EM wave is a propagating disturbance in the existing electromagnetic ...

0

The electric field is given by $$E=-\nabla V-\frac{\partial A}{\partial t},$$ where $V$ is the electric potential and $A$ is the vector potential. So you can get the electric field from the potential. Or if you have measurements of a large enough set of points that are close together compared to the scales over which the field varies you might be able to ...

0

The electromagnetic field is mediated by the exchange of virtual photons. This is described by Quantum electrodynamics. It's this exchange that bridges the gap of empty space. I think the answer to this question has a pretty good representation of how an electromagnetic field propagates (at the speed of light) in vacuum.

0

Field lines are tangential lines obeying the Gaussian law. The can only meet at an angle at discontinuities in the field like point charges or surfaces with inhomogenous magnetic flux and similar.

2

Field lines are a visual representation of a mathematical construct, like a graph of a function. The defining properties of this visual representation are The field lines run parallel to the field at every point. The density of the field lines in an area is proportional to the strength of the field. The second property tells you that the field lines can ...

4

Electric field lines reveal information about the direction (and the strength) of an electric field within a region of space. If the lines cross each other at a given location, then there must be two distinctly different values of electric field with their own individual direction at that given location. This could never be the case. Every single location in ...

13

Electric field lines are a visualization of the electrical vector field. At each point, the direction (tangent) of the field line is in the direction of the electric field. At each point in space (in the absence of any charge), the electric field has a single direction, whereas crossing field lines would somehow indicate the electric field pointing in two ...

4

The electric field at any point is the sum of all the fields due to each individual charge in the system. The field has a magnitude and a direction. The field lines are a representation of the magnitude and direction of the field over an illustrated area. The field lines point in the direction of the field. If lines from two sources were to cross, we could ...

1

If you take a permanent magnet, and place a sheet of paper over it. Now sprinkle iron filings on it, and you pretty much get this diagram. This has been the mainstay of field theory since Faraday's time. A test charge at rest will begin to move in the direction of the field line. Since there is nowhere that it can rest where there is more than one ...

0

At any point the electric field is the vector sum of the fields from the two charges. So while the fields from $A$ and $B$ are indeed in opposite directions at your point $p$ you just add them (well, subtract their magnitudes since they're in opposite directions) and this gives you the net field. I wouldn't take the field lines too seriously. They are not ...

1

The solution of Laplace's equation, $\nabla^2 \phi =0$, is a harmonic function, which has the property that it has no local minima or maxima. This implies that $\vec{E} = -\vec{\nabla}\phi$ can not be zero if $\phi$ is not constant, hence it can be used to define a curve, the field curve with tangent vector $\vec{E}$ pointing in the direction that $\phi$ ...

0

A different route to the same result, which you may or may not find more intuitive, would be By definition the electrostatic field is the sum of the Coulomb fields of all of the source charges. (In case of a continuous charge distribution, we can either consider the limit of a collection of ever smaller point charges, or replace the sum with an integral, ...

3

A force is said to be conservative if its work along a trajectory to go from a point $A$ to a point $B$ is equal to the difference $U(A)-U(B)$ where $U$ is a function called potential energy. This implies that if $A \equiv B$ then there is no change in potential energy. This fact is independent of the increase or not of the kinetic energy. If a conservative ...

0

If electric field lines would have been closed loops, then there would have been no isolated electric charge as like there exists no isolated magnetic pole. So, this is another reason why electric field lines can't form closed loops. The magnetic field lines of a magnet form continuous closed loops, this is unlike electric dipole where the field lines ...

0

Robin is right in stating that if Electric Fields form closed loops, they wouldn't be conservative. But keep in mind that non-conservative Electric Fields can also be produced in some situations, like changing magnetic flux.

11

If there was a closed field line a particle following that line would eventually return to the same place but having a different energy so the field would not be conservative.

1

Dipole $\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$ $\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$. \begin{align} \ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac ...

Top 50 recent answers are included