# Tag Info

36

Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but this is currently the most fundamental explanation known. It's ...

12

The maximum charge a capacitor stores depends on the voltage $V_0$ you've used to charge it according to the formula: $$Q_0=CV_0$$ However, a real capacitor will only work for voltages up to the breakdown voltage of the dielectric medium in the capacitor. So in reality, for every capacitor there is a maximum possible charge $Q_{max}$ given by: $$... 10 I suppose you mean k_e=\frac1{4\pi\epsilon_0}. That comes from the fact that Coulomb's law can be stated as :$$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2 $$Now, \epsilon_0 is the electric constant, or the permittivity of free space, and it essentially scales the force. The 4\pi r^2 comes from the surface ... 9 Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will". If you take q|\vec v| instead of q\vec v, you get something related to current (generally current times a factor). Current is conserved at a junction. Regarding your equal-and-opposite situation, the closest ... 9 If you want to avoid factors of \pi in the more fundamental equations like \nabla . E = \rho / \epsilon_0, you have to accept them where they belong, for instance in: E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}. As remarked by others, Newton failed to put a factor 4 \pi into his gravitation equation (he stipulated g = G \frac{M}{r^2}, instead of ... 9 James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential \Phi relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is:$$\Phi_{inner} = \Phi \,q\, ...

8

Of course you can define such a quantity, but the question is: does it mean anything physically? Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great ...

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 ...

7

+1, Good question,. While I don't think your idea has much of a physical implications, it is a good analogy (in my opinion, at least). A fair approximation to General Relativity is Newtonian Gravity. A better one is Newtonian Gravity with some special relativistic corrections (I mean a modification to Newton's gravity where the masses $m$ are replaced ...

7

The mistake you made is in the way you stated Coloumb's law. It's either $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}}$$ OR $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}}$$ but definitely NOT $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$

7

Freely-moving charges placed on a line will tend to fly away from each other - with no equilibrium position possible - unless there is some potential that confines them to a specific region. Enforcing the charges to lie within an interval $[0,L]$ will always mean one charge is at either end, so you might as well consider $n-2$ charges confined by the ...

7

This problem has been solved by Griffiths in Charge density of a conducting needle. David J. Griffiths and Ye Li. Am. J. Phys. 64 no. 6 (1996), p. 706. PDF from colorado.edu. The problem is nontrivial.

7

This is a more down-to-earth answer as opposed to the fancy mathematics in the other one. This problem is easily solved numerically. The equations are easily stated: inverse-square forces to the right from the particles to the left and to the left from the particles to the right. Thus, for a system of $n+2$ charges where the first and last are fixed at $x=0$ ...

7

Does the fact that every rain drop falls in their respective straight lines all parallel to one another imply that those lines are physically real? No. It is just the tendency of gravity to act between two massive objects--a straight line is simply the least inaccurate way to describe this interaction. You can also draw additional curved lines linking the ...

6

Four possibilities come to mind, in decreasing order of feasibility: Is barefoot an option? I'm willing to bet it will significantly mitigate the buildup of charge. The two of you only experience a shock on contact because you are at different electric potentials. If you can't keep him at your potential, why not try to join his? Before helping him down, ...

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 ...

6

I usually find it easier to use model multipoles that are surface charges on a sphere, rather than point charges on some polyhedron's vertices. These charge densities are given in general by $$\sigma_{lm}(\theta,\phi)=N\cos(m\phi)P_l^m(\cos(\theta)).$$ Thus a monopole is constant, a dipole has $\sigma=\cos(\theta)=z/r$, a quadrupole has the form ...

6

I can give you an intuitive view from a physicist. Charges are the sources and sinks for the electrical field. Consider the extreme case where the volume enclosed by the surface is empty space, so no charges. Then any field line that enters the volume must exit the volume somewhere else. Thus, the integral of the field over the entire surface is 0. If ...

6

It is a misconception to think that just because the 29th electron is outside the gaussian surface, it will not have an effect on the electric field inside it. The total flux through the surface is indeed zero, but that doesn't mean there is no influence: imagine a point charge and draw up a sphere next to it. The electric field goes in on one face and out ...

5

If it is in air (or any other substance), there is a limit where the electric field of the object is going to be enough to ionize the surrounding medium, and the resulting current will drain the object of its charge. Similarly, if the object is immersed in vacuum, you will eventually have an electric field sufficient to "polarize the vacuum" by creating ...

5

Using the method of images, you can calculate the force between the ring of charge and the sphere. Assume the sphere is on the z axis with it's center on the point $z$, a radius of $R_s$ and the ring's radius is $R_r$ with a charge density $\lambda$. So $z$ denotes the center of the sphere. To calculate the force, you can replace the sphere with a charged ...

5

There is a standard book which contains everything about electrostatics, the Laplace/Poisson equation and boundary conditions: Classical Electrodynamics by J. D. Jackson. Get the book from the library of your choice, read all chapters labeled "Electrostatics", and you will find the answers to all your questions (if you are simulating this, you need to know ...

5

Let me first comment that the statement electric fields cancel while the electric potentials just add up algebraically is not actually correct. Electric fields add due to the principle of superposition (see the section on superposition in the wikipedia article). However, when two electric field vectors are of the same magnitude but point in ...

5

Since this is a homework-type problem, here are some Hints for the force The electrostatic force $d\vec F$ on a small segment $dl$ of the rod given the field $\vec E$ of the other rod is $$d\vec F = \lambda\, dl \,\vec E$$ Determine the field of one rod, and use the above expression to integrate the force it exerts on the other rod. This is a 2D ...

5

First of all note that $k$ is not dimensionless, it is $k = \frac{1}{4 \pi \varepsilon_0}$, and $\varepsilon_0$ has dimensions of $\frac{ \text C^2}{ \text {N m}^2}$. So you have already $\frac{ \text{V C}^2 \text m^2}{ \text {N m m} ^2}$. Also, volt can be expanded as $\text V = \frac{ \text {N m}}{ \text C}$, so one gets $$\frac{ \text C^2}{ \text {N ... 5 The exact solution is$${\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}}.$$Clearly the field inside the conductor (that is, for r<R) vanishes. Here Q is the total charge on the disk. The field, for large values of r, looks ... 5 First off, what you describe only happens for highly amplified speakers. The current change when you touch the wires is pretty tiny and you'll only hear a sound when that signal is being amplified significantly before being sent to the speakers. There are many reasons why a tiny bit of electric current flows from your body to the speaker wire so I'll only ... 5 It's not intuition.It's a problem which can be solved. First we identify the sign of the charges. By seeing the direction of field lines we can see that the sign of charges. Field lines originate from +ve and end at -ve charges. Next by Definition of Flux, The number of field lines cutting per unit surface surface . And Gauss' Law The flux ... 5 The physical reason for the appearance of a 4\pi somewhere in the theory is the spherical symmetry of the problem and is discussed more in other answers . Here I want to quote an interesting argument from Arnold Sommerfeld's Lectures on Theoretical Physics Vol III, which has a section dedicated to this issue. If you remove the 4\pi from the force law ... 5 Let's suppose we have a source particle and and a test particle a distance r from each other. Upon measuring the force on the test particle, we find some value F_\text{one source}. By varying the distance, we discover it depends on the inverse distance squared:$$F_\text{one source}=\frac{C}{r^2}, where $C$ is just some constant. (No 'charge' appears ...

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