Hot answers tagged electrostatics
7
In an attempt to be brief: The big thing to remember is that the flux is also proportional to the area (technically, the surface integral of the field over the area). Crudely speaking, the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases ...
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
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 ...
3
Not always. All of your Gaussian surface should be in a linear dielectric with constant electric permittivity $\epsilon$ to be able to use gauss law and derive that formula. With this conditions it's true most of the times.
Here you can use again the gauss law:
$\vec D = {Q_a \over 4 \pi r^2} \hat r$
But we know that for linear dielectrics: $\vec D = ...
3
Edge effects. After the electron leaves the capacitor, the electric field winds up slowing it back down.
Let's assume the capacitor is infinitely-massive and that the acceleration of the electron is small enough that we can ignore radiation.
Then if you were to idealize the electric field of the capacitor, treating it as a uniform field between the plates ...
3
Consider this as an appendix to other answers. I hope that this will you to believe that 'the number of lines' is not only a qualitative description of the flux, but also quantitative.
Actually, with the use of Gauss' law, you can prove that the flux through a surface is proportional to the number of lines crossing it (under certain conditions). ...
3
Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".
If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.
Regarding your equal-and-opposite situation, the closest ...
3
First off, what you describe only happens for highly amplified speakers. The current change when you touch the wires is pretty tiny and you'll only hear a sound when that signal is being amplified significantly before being sent to the speakers.
There are many reasons why a tiny bit of electric current flows from your body to the speaker wire so I'll only ...
3
Our current understanding suggests that black holes can have electric charge, and that in addition to mass and angular momentum, these are the only ways that black holes can have distinguishable physical properties. This is a result of the famous no-hair theorem, although keep in mind that no definitive proof of this theorem yet exists.
As this wikipedia ...
3
Maxwell's equation $\nabla \cdot \mathbf{E} = 0$ only states that the electic field does not change in a region devoid of charge. The assumption $\mathbf E = 0$ states that the electric field actually vanishes.
For example between the plates of a condensator the region is devoid of charge and the electric field is constant but non-zero. Such a behaviour is ...
3
When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes ...
2
These are two separate questions. It's better if you don't try to combine two questions into one.
In answer to the first question, yes, a black hole can have a measurable charge. You measure it the same way you'd measure any other charge. This is all purely theoretical, however. Many real-life black holes have been observed and characterized by their ...
2
On the upper end, Coulomb's law has not been observed to break for any large collection of charge that can be put together. In principle, if you tried to put more and more charge together then there would be a lot of energy stored in the field, and if the mass equivalent of this energy density got too high, there would be general relativistic effects to ...
2
All the answers are great. I'll just add here a more quantitative explanation, though an easy one IMHO, that I really liked, from the book "Fundamentals of Physics, Volume I" by B.M. Yavorsky and A.A. Pinsky:
Both of the tiny areas $\Delta S_1$ and $\Delta S_2$ are sharing the same solid angle $\Delta \Omega$, thus, the absolute value of the flux is the ...
2
When we're calculating the energy stored in a capacitor we normally assume it is isolated i.e. there are no other charges nearby to affect it. This makes the calculation nice and simple: the energy is proportional to $Q^2$ and the energy is stored in the electric field around the capacitor.
However in your question you are introducing another charge, your ...
2
The energy is of course coming from the electric field of the capacitor. The energy of any capacitor is always stored in it's electric field. If an electron is initially positioned very far away and then moves close to the capacitor, it's being pulled by the field and that means energy is being transferred. The electric field get's a little weaker - loosing ...
2
Implicit in the derivation of $E=\rho/2\epsilon_0$ (you were off by a factor of two by the way) is the assumption that the charge can be approximated by a continuous charge distribution. You correctly note that as you approach a real sheet of charge composed of point charges this treatment breaks down. Let's calculate the $z$-component of the electric field ...
2
New version
The problem in your demonstration is when you write down $\vec{A}\cdot\vec{B} = ||\vec{A}||\,||\vec{B}||\,\cos\theta$. More exactly, in your case $||d\vec{r}||\neq dr$ because $dr<0$ when you go from $\infty$ to $r$ and a norm is positive by definition. So the sign error is introduced from 3rd to 4th line.
Old version
The demonstration on ...
2
As far as I can tell, this "argument" is not a meaningful derivation of $\mathbf E = -\nabla \Phi$. In step 4, one seems to assume that $Edx = -d\phi$ which is essentially what one is trying to prove in the first place, so the argument seems circular.
The precise derivation of this fact relies on the crucial fact that the electric field is a conservative ...
2
The electric field is a conservative vector field which implies that there exists a function $V$ for which
$$
\mathbf E = -\nabla V
$$
We call this function $V$ the electric potential. There is no mathematical need to first define potential energy. One can then physically interpret $V$ in terms of a "potential landscape" to get intuition for what it ...
2
When you calculate work, you do so along a given path. Here, that path has tangent vector $d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward.
Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate $r$ to parameterize the path, in ...
2
I'll suppose your disk (radius $R$) is uniformly charged with surfacic charge density $\sigma$ (your disk being conductor, any charged Q deposed on it will be distributed all over it so that $Q=\sigma\times\pi R^2$). The field outside the disk is necessarily radial so we can place the point $M$ where we would like to compute the field on the $x$ axis so that ...
2
Your answer would have been correct if, for example, the spheres were non-conducting and if the charges were distributed uniformly over their surfaces.
However, since the spheres are conducting, the surface charge distribution on each sphere will be altered because of the repulsion from the charges on the other sphere. In particular, the charges on each ...
1
There are many materials that can be charged by triboelectric effect. Tipicaly you can observe this effect rubbing a material like wool and amber.
The phenomenon is quite complex but it's in great part because the different electron affinity of the materials (one loses easily an electron and the other captures an electron).
1
I think every fundamental definition is kind of going in circle.
I would say an electric charge is something that obeys Maxwell's laws.
But to write those laws, you have to know $\vec{E}$ and $\vec{B}$ which need a definition of an electric charge.
At the end you just group things that look/react alike and named them.
The problem arise when you have to ...
1
Practically, for a macroscopic body such as a chunk of metal, the charge on that body is the difference between the number of electrons and protons in the body. It is hard to knock a proton out (can be done, though), but for conductors we can push and pull electrons out by supplying a bit of energy (back to that in a moment). However, instead of saying the ...
1
Since you are a high school student I'm going to make this as simple as possible. This should be read in addition to joshphysics's (correct) answer. The purpose of my own answer is to specifically fill in the steps that the book left out, but which are nevertheless correct.
First, I will take a guess that you have seen basic integral calculus and that your ...
1
$\mathbf{r}$ is a position vector and $\mathbf{s}$ is a displacement vector between two points, let say A and B. In general case, they are not equal, but they can be if we properly choose the origin of the coordinate system: A={0,0,0} or B={0,0,0} The sign depends on at which point A or B the origin is placed.
1
Your mistake is that in the expression
$$
Q=CV
$$
the symbol $V$ here represents the magnitude of the potential difference between the spheres. Thus, since $b>a$ here, you need to switch the order of $b$ and $a$ in the first expression you wrote down for $V$ if you want to plug it into the expression defining capacitance (in other words, you need to ...
1
Just to be clear, the potential energy of a particle of charge $q_2$ at a distance $r$ from a source of potential (supposidely at zero) of charge $q_1$ is the work that an external operator has to provide to bring the particle from infinity to $r$ at constant velocity.
This reads then:
$\int_{\infty}^r \vec{F}_{op}\cdot \vec{ds}$
As people have said, the ...
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