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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 ...


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 ...


10

Although this is quite an old question, I have to disagree with answer by LuboŇ°. First, Pauli exclusion principle says that no two fermions can share the same state. But, if the electrons have different spins (i.e. are in so called spin-singlet state), then they can be in the same positional state. Next, indeed, in classical case, two charged particles ...


10

Coulomb's law becomes invalid at distances of the order of the electron Compton wavelength and smaller, due to vacuum polarization. To first order in the fine structure constant, the electric potential due to a charge q at the origin is given by: $$V(r) = \frac{q u(r)}{r}$$ where $$u(r) = 1 +\frac{2\alpha}{3\pi}\int_1^{\infty}du ...


10

This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law \begin{equation} \nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0} \end{equation} If we integrate this over some volume $V$ and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} ...


8

You have to realize that the system is invariant under rotations about the normal to the plane. Then then electric field must also be invariant under these rotations. An electric field component in the plane does change under such a rotation, so such a component must not exist if we have this invariance. Thus the electric field is purely along the normal to ...


8

The answer by @NowIGetToLearnWhatAHeadIs is correct. It's worth learning the language used therein to help with your future studies. But as a primer, here's a simplified explanation. Start with your charge distribution and a "guess" for the direction of the electric field. As you can see, I made the guess have a component upward. We'll see shortly why ...


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

It doesn't hold for arbitrary shapes. The reason it works for spheres is that when you have a spherical charge distribution and a concentric spherical Gaussian surface, the whole system is invariant under rotations around the center of the spheres. If the electric field were different at different points on the Gaussian sphere, you could rotate the whole ...


6

I) Right, the differential form of Gauss's law $$\tag{1} {\bf\nabla} \cdot{\bf E}~=~ \frac{\rho}{\varepsilon_0} $$ uses the relatively advanced mathematical concept of Dirac delta distributions in case of point charges $$\tag{2} \rho({\bf r})~=~\sum_{i=1}^n q_i\delta^3({\bf r}-{\bf r}_i).$$ Note in particular, that it is technically wrong to claim (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 ...


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

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

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

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

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

for an infinite sheet of constant charge, the text says that the electric field is constant on any one side of the sheet. But that seems intuitively wrong to me, since I would think the field should be stronger the closer a point is to the sheet. There's a geometric scaling argument at hand, and you probably need to appreciate Gauss' Law to get a real ...


5

The only property of metals used in deriving $C=\varepsilon A/d$ is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter. But if you use something other than metal, then it will of course change the capacitance.


5

In physics and engineering, we often abstract and idealize a physical problem to gain insight into the physics, e.g., infinite plane of charge, infinite line of charge, point charge, etc. Now, it goes without saying that if these idealizations didn't represent good approximations of relevant physical systems, they wouldn't be used. With regards to your ...


5

The integrand $\vec E \cdot d\vec r$ is $E\,dr$, not $-E\,dr$. The evaluation of the dot product is sort of done for you when you specify the curve on which you are integrating (i.e., your limits of integration in this case). You've double-accounted for the relative directions of $\vec E$ and $d\vec r$. I suspect the underlying confusion is that you are ...


5

Take a look at the conventional form of Maxwell equations. They tell us that Gauss's law actually applies every time. However, to get the field $\vec{E}$ from the charge distribution by the usual methods, we also need to know that $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} = 0$$ Because otherwise the field could not be generated by the ...


5

$$A=\pi r^2$$ $$\frac{dA}{dr}=\pi\cdot2r$$ $$dA=2\pi rdr$$ Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$ You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit ...



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