Hot answers tagged electric-field
12
1) yes, it basically will find a non-optimal solution. At every point, the top of the ray looks for the bigger potential gradient, the charge in the surrounding volume grows, polarizing surrounding material (air, in this case) until a bigger gradient shows up and the ray continues over that direction. This is why the lightining path looks like a jigsaw; its ...
9
Electric current, by definition, is a flow of charged particles.
When someone says it is the propagation of the electric field, usually he means the following:
The velocity of the electrons in the wires is very slow (few cm/s if I remember it right), but when one turn on the light he doesn't see any delay. The lamp starts lighting when the electrons start ...
7
There's an old and clichéd but actually pretty good analogy for understanding electric circuits, and that's to think of the circuit as water plowing through pipes.
If you have a pipe full of water, and you turn on the tap at one end, water immediately starts flowing out of the other end. Well not quite immediately: when you turn on the tap you raise the ...
7
Be careful here.
Gauss's law tells you that the flux through the (whole) closed surface is proportional to the enclosed charge and therefore zero in this case.
That's one fact.
The second fact is that you have a constant electric field in this region of space, and that means that the flux through the circular end-cap (which is not a closed surface) is ...
6
No, the statement is false even in the electric case. At the very beginning, the acceleration is $\vec a \sim \vec E$ so they have the same direction at $t=0$: the tangents agree.
However, as soon as the particle reaches some nonzero velocity $\vec v \neq 0$, its acceleration is still $\vec a\sim \vec E$, in the direction of the field lines, however its ...
6
Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates.
First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the $z$-direction. I will follow the notation of the Wiki article:
$$
x=a\cosh\mu\cos\nu\cos\phi\\
y=a\cosh\mu\cos\nu\sin\phi\\
z=a\sinh\mu\sin\nu
$$
where ...
5
I'll give you the derivation from my book which includes a nice way to see how the delta functions arise:
..............................................................................................................................................................
We can derive the potential field $\vec{A}$ and the electromagnetic fields $\vec{E}$ and ...
5
You tell us that one surface of the box is at $B$, but you're a little vague on where the opposite face is. You do say that your surface is "between the two sheets", so I think you may mean that the surface is entirely contained in the space between the two sheets. The box does not intersect any charged surface. With that, and a uniform electric field in ...
5
Gauss' law is applicable for a finite wire. But, it's useless in this case.
In the infinite example, you assume some things due to symmetry, namely:
It's pretty obvious why these things can be assumed--moving up and down the wire should not change $\vec E$, so we take it constant. Also, there should be no direction bias, so $\vec E$ has no component ...
5
Electrons will flow against the electric field lines because their charge is negative, and the electric field thus exerts a force $\mathbf{F}=q\mathbf{E}$ on them which is in the opposite direction. Thus electric field lines inside the wire go from the positive to the negative terminal and the electron flow goes from the negative to the positive terminal. ...
5
The wikipedia article is quite good on this subject.
For any discharge in the air the molecules of the air must be ionized. This ionization happens during thunderstorms because of the high static electric fields carried by the clouds which generate "streamers", i.e. paths for the electrons to flow downwards. Corresponding streamers are formed by conductors ...
5
The exact solution is $${\bf E}(R<r, \theta =\pi/2)=\frac{Q}{4 \pi \epsilon_0 }\left(\frac{1}{r^2}\right)\sum_{l=0}^{\infty}\frac{(2l)!}{2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l}\hat{{\bf r}}.$$
Clearly the field inside the conductor (that is, for $r<R$) vanishes. Here $Q$ is the total charge on the disk. The field, for large values of $r$, looks ...
4
The OP wants an intuitive answer to an intuitive obstacle to seeing its truth. Well, the intensity of the flux is like how many lines we draw per unit area. No one line « loses strength » so to speak. (There is no dissipation, no friction.) If it is a point source, the lines are not parallel, they diverge, and the greater distance between the lines leads ...
4
It is indeed correct that only the difference between two potential energies is physically meaningful. An in-depth explanation follows. For the rest of this answer, forget everything you know about potential energy.
I suppose you know that when you have a conservative force $\vec{F}$ acting on an object to move it from an initial point $\vec{x}_i$ to a ...
4
No, there is nothing special about your right hand as compared to your left one. (Well, there might be, if you're a baseball player or a fiance, but there's nothing in classical electromagnetism that makes it special.)
If you set up two wires next to each other and run current through them, they will attract if the current runs the same way and repel if ...
4
Yes, your friend is right. Within electrostatics, an electric field $\vec{E}$ should be curl-free $\vec{\nabla} \times\vec{E}= \vec{0}$. The drawn electric field lines looks like the electric field is of the form
$$ E_x=E_x(y), \qquad E_y=0, \qquad E_z=0, $$
cf. the rule that to depict the magnitude $|\vec{E}|$, a selection of field lines is drawn such ...
4
If you followed the arguments carefully and checked what is demonstrably right and what is not, you would agree that what the argument actually does is to prove that a uniform electric charge density cannot have a uniform electric field. Your original task was to solve Maxwell's equations (well, Gauss's law), so if you find out that the equations aren't ...
4
(as per Chris White's suggestion)
The diagram is confusing. It is drawing two sets of field lines: one set due to plate A (as if plate B didn't exist) and another due to plate B (as if plate A didn't exist). It is not showing the total field. This doesn't represent the total field if both plates are present!
The electric field is a vector field $\vec{E}$: ...
4
The field strength is the negative first derivative of the potential.
For example, in Cartesian coordinates, with electric potential $V = V(x,y,z)$, the electric field is:
$$
\boldsymbol E =-\nabla V= -\frac{\partial V}{\partial x}\boldsymbol i - \frac{\partial V}{\partial y}\boldsymbol j - \frac{\partial V}{\partial z}\boldsymbol k
$$
Note that you can ...
4
I'd like to add a bit of mathematical detail the (correct) statements by DJBunk. Let a scalar function $f$ be given (let's not restrict ourselves to the electric potential). For any unit vector $\mathbf n$, we can define the directional derivative $D_\mathbf{n}$ of the function $f$ in the direction $\mathbf n$ as follows:
$$
D_\mathbf{n}f(\mathbf x) = ...
4
The electric field of a negative point charge points towards the point charge as a result of the definition of the electric field of a point charge. To see this, recall that the electric field of a point charge $q$ is defined as
$$
\mathbf E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\mathbf e_r
$$
where, $r$ is the distance to the charge, and $\mathbf e_r$ ...
4
Although you say that there are great similarities in gravitational and electric fields, there is one thing that is fundamentally different in the two - that gravitational field produced by a mass is always in the same direction(towards the mass) while in case of electric(electrostatic) fields two different kinds of electric charges produce fields in ...
3
The simplest explanation I know of requires only one test charge and two reference frames with a relative velocity between them.
Frame 1: The charge is at rest. It is the source of a (purely) electric field.
Frame 2: The charge is moving. It is a current, and the source of a magnetic field.
3
Conductors are defined by the freedom of some of the charges inside to more with little resistance.
So, if there were a non-zero field, what happens? Answer: some of the free charges move until the field is again zero.
You might be wondering if there are limits to this claim, but a introductory book of that sort is not worrying about extreme situations. ...
3
The "first level" answer was given by nibot in a comment.
The entire conductor must be equipotential. If there were a potential difference from one part of a conductor to another, free electrons would move under the influence of that potential difference to cancel it out.
However, since I have similar curiosity myself I'm going to try to answer in ...
3
The factor of two is coming from the place you identified.
Think about throwing out that factor of two, so you're considering only the bottom hemisphere. When you make your Gaussian shell and have it enclose charge in the bottom hemisphere only, the charge is no longer uniformly distributed inside your Gaussian shell. Thus, the electric field created by the ...
3
As it happens I've just been reading "17 Equations That Changed the World" by Ian Stewart and he gives the derivation. I strongly recommend the book, but if you just want the derivation you can find it on Wikipedia. Since we're not supposed to just give links I'll copy the stuff from Wikipedia here:
Start with Maxwell's equations:
$$ \nabla \cdot ...
3
It goes out forever, but the total energy it imparts is finite. The reason is that when things fall off as the square of the distance, the sum is finite. For example:
$$ \sum_n {1\over n^2} = {1\over 1} + {1\over 4} + {1\over 9} + {1\over 16} + {1\over 25} + ... = {\pi^2\over 6} $$
This sum has a finite limit. Likewise the total energy you gain from moving ...
3
The plot you're trying to imitate seems to not go to infinity. I'd suggest you play around with potentials of the form $$V=\frac{V_0}{\sqrt{x^2+y^2+\delta^2}}.$$
E.g. Something like
Plot3D[-(1/Sqrt[x^2 + y^2 + \[Delta]^2]) /. {\[Delta] -> 1/10}, {x, -1, 1}, {y, -1,
1}, PlotRange -> Full, AspectRatio -> 1, ColorFunction -> Hue]
gives
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