Tag Info

Hot answers tagged

17

Gauss's law is always fine. It is one of the tenets of electromagnetism, as one of Maxwell's equations, and as far as we can tell they always agree with experiment. The problem you've uncovered is simply that "a uniform charge density of infinite extent" is not actually physically possible, and it turns out that (i) it is not possible to express it as the ...


17

The force does not change instantaneously, the correct way the electromagnetic field of (and thus the force exerted by) a moving electric charge is given by the Liénard-Wiechert potential, where one can see that the effect of the charge does not travel faster than light.


17

To add to ACuriousMind's answer on the Liénard-Weichert potentials, you can put these formulas into an even more wonderfully descriptive form since you can derive Feynman's formula from them for the radiation from a moving charge: $$\vec{E} = ...


16

You smell ozone ($\mathrm{O_3}$, from the Greek word ozein for "smell"), and maybe nitrous oxide - the reaction product of oxygen and $\mathrm{N_2}$. There is a nice description of the formation and action of ozone at this link. Briefly: Oxygen molecules ($\mathrm{O_2}$) can be dissociated (broken into atoms or ions) by either UV light, or electrical ...


11

You want a gas so you don't need to expend energy vaporising the propellant. You also want the gas to be as dense as possible so you can get as much impulse per unit volume of propellant as possible. It's also nice if the gas is inert and non-corrosive so you don't need to worry about it degrading or corroding whatever you're storing it in. Finally it's nice ...


11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.


11

The statement "electric field inside a conductor is zero" is true only after charges have distributed themselves in the most optimal way on the surface - it is an electrostatic result. Starting with an arbitrary charge distribution, there will be forces that cause a redistribution of the charge until, for a sphere, they are distributed uniformly. At that ...


11

You are right - potential is a scalar. But a dipole moment is a vector - it has magnitude and direction. When sodium channels open up, charge flows. Lots of charge moving a little bit causes a change in the dipole moment of the heart. This in turn induces charge to move elsewhere in response (the dielectric properties of tissue cause a propagation of the ...


9

What the picture shows is a corona discharge (see also Wikipedia). It isn't a circuit in the usual sense of the word. It happens because the voltage is so high that it raises the electron energy to above the work function and the electrons just leak off. In effect the coil is charging the air around it. The charge will end up on the furniture, walls, floor, ...


8

You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner ...


8

This may help your, it comes from Rutherford scattering by which they determined that the atom has a hard core. It is positive alphas against positive nucleus, but the math is the same. Determining the closest approach to the nucleus amounts to calculating the minimum distance for the hyperbolic orbit which is produced by the coulomb repulsive force. ...


8

There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy $W$ of a uniform spherical charge distribution of radius $a$ and total charge $Q$ $$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$ Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ...


8

Ideally, test charge should not affect the charge distribution of the source. An infinitesimal charge will ensure, for example, that the electric field it produces does not redistribute charges on any conductors in your system. A large test charge would polarize nearby objects, thus affecting the field you're trying to measure in the first place.


8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...


7

The physical meaning of the capacitance is precisely given by $\mathrm{d}Q=C\cdot \mathrm{d}V$: $C$ tells you how much charge there will be in the capacitor per voltage applied. For all capacitors, the linearity holds fairly well. Generally speaking, capacitance is given by a Q-V curve, which may consist of a linear region, a saturation region and a ...


7

If you are talking about point charges then, as explained above, the answer is no. But in the case of non-uniform charge distributions, it is possible for same-charge particles to attract, if they are sufficiently close. As an example, the following two particles are identical, each having a net charge of -1. Plotted below them is their potential energy as ...


6

The vanishing of closed line integrals means that the field is conservative. Since $\oint \vec E \cdot \mathrm{d}\vec l$ is equivalent to $\vec \nabla \times \vec E = 0$, the "physical interpretation" is the the electric field is irrotational, i.e. it has no "vortices". The, more valuable, mathematical implication is that there is a scalar potential whose ...


6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...


6

The reason is the same as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) until they don't feel a force any more. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


6

Electrostatic refers to the case where the fields are not time dependent. In that case the Maxwell's equations reduce to: $$\nabla \cdot E =\frac{\rho}{\epsilon_o} \\ \nabla \times E = 0 \implies E=-\nabla \phi \\ \text{then,} \nabla \cdot \nabla \phi = \nabla^2 \phi = -\frac{\rho}{\epsilon_o} $$ The solution to the last equation is: $$ \phi = ...


5

Now, the constants C1,C2,C3 appearing when we separate variables on Laplace's equation for electrostatic potential has some physical meaning? If they do, what is it? The constants are the related to the square of the spatial (angular) frequency or a spatial growth/decay constant. For an example of spatial frequency, let $$X(x) = A \sin (k_xx) + B ...


5

So in the first case, when talk about a plain circular ring, I assume you mean an annular ring, with a well defined inner radius and a different well defined out radius. With a positive charge at the center of the annular ring, positive charges will be repelled outward and negative charges attracted inward. Incidentally, not all the positive charge will go ...


5

For $H\gg R,L$ And for $L\gg R,H$ you get pretty much the same thing. First off, $(H+L)^2\sim H^2$ and the same goes for $(H-L)^2$. That means that $(H+L)^2+R^2\approx (H+L)^2$. However, $(H+L)\not\approx H$, which means that $\sqrt{R^2+(H+L)^2}\approx H+L$. This makes the first approximation have $\frac{1}{H+L}-\frac{1}{H-L}$ in it. The second ...


5

I want to complete the other answer by addressing the difference of a charge distribution made statistically up by a huge number of electrons , and what an electron means: At the level of elementary particles, one of which is the electron, there are no charge distributions, as elementary particles are point particles, and charge is a quantized quantity ...


5

Noble gases have the advantage of being chemically inert, so that they are less likely to react with atoms in the electrostatic grids. Since ion thruster to date have been deployed only on unmanned transports, regular maintainance is not an option. Because of that, Noble gases are favoured over, say, hydrogen One reason to pick Xenon over Argon though, ...


5

Can anyone provide me with a physical interpretation For an electric field satisfying the equation, the work associated with (slowly) moving a test charge around a closed path is zero. To see this, recall that the electric force on a charge is $$\vec F = q\vec E $$ The work associated with moving a particle along a closed path is $$W = \oint \vec ...


5

There's no minimum distance. Yet, as the two particles get closer to each other, they will either scatter off each other (in the quantum mechanical sense of interacting via Feynman diagrams) or form a bound system - if we're talking electron-positron (which is as close to point charges as it gets), they might become positronium, but that won't last long, ...


5

The electric potential $\phi:\mathbb{R}^3\to\mathbb{R}$ is the solution to Laplace's equation and therefore a harmonic function. Harmonic functions enjoy several nice properties, some of them listed on the Wikipedia page. Concerning OP's second point, let us mention that there is a theorem similar to Liouville's theorem from complex analysis that a bounded ...


5

The problem here is that you've failed to specify a boundary condition. Consider an electrostatics problem where you're given a charge distribution $\rho(\mathbf{r})$ and asked to find the electric field $\mathbf{E}(\mathbf{r})$. The electric field is the solution to the set of differential equations $\nabla \times \mathbf{E} = 0$ and $\nabla \cdot ...



Only top voted, non community-wiki answers of a minimum length are eligible