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19

Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law). You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or ...


14

This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below). Even worse, gravity and electricity don't even follow this in ...


12

This is a good example of a procedure that happens in many areas of physics. In general, physical laws - and particularly conservation laws - tend to be most naturally phrased in integral form, or even in mixed integro-differential form. For an example of the latter, consider the integral form of Faraday's law: $$ \oint_{\partial S}\mathbf{E}\cdot\text ...


11

If the Gauss rifle shoots a projectile with exit speed of $v_1$ and mass $m_1$, then its momentum will be: $p=m_1v_1$. Because of momentum conservation law, the rifle will have the same momentum in opposite direction. If the rifles mass is $m_2$, the rifle will start moving in the opposite direction with end speed of: $v_2 = \frac{m_1 v_1}{m_2}$. But, as ...


9

Richard Feynman explains it wonderfully in his lectures on the "Character of Physical Law". You could have a look at this video. One way to see it is that the inverse square law just is. There were experiments done to determine how one mass affects the other and then the result was that the force between them varies as the inverse square of the distance ...


8

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 ...


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

If I understood your question correctly, then you want a simple experiment to demonstrate that magnetic monopoles cannot exist. The simplest way to explain this to a high schooler would be to actually break a small piece of magnet, and then make the student realize that the poles of the magnet haven't been 'split'; instead, both the pieces contain two poles. ...


6

I would say yes ! Actually some theories explaining quantum gravity use also this reasoning: gravity is a very weak interaction at a quantum level because it "leaks" into other dimensions, not observable at our scale, but that are present at this scale. The mathematical tools are different, but if you just think about gauss's law you can imagine one ...


6

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 ...


6

In the equations as you've written them, the constant of proportionality is an outward-pointing vector for the electric field and an inward-pointing vector for the gravitational field. Or in other words, if you take the radial component only: it's a positive constant for the electric force and a negative constant for the gravitational force. The details: ...


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 ...


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

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 ...


5

Gauss' law and Coulomb's law are equivalent - meaning that they are one and the same thing. Either one of them can be derived from the other. The rigorous derivations can be found in any of the electrodynamics textbooks, for eg., Jackson. For eg., consider a point charge q. As per Coulomb's law, the electric field produced by it is given by $$\vec{E} = ...


5

Gauss's law would not be valid. You can imagine the electric field "flowing out" from positive charges and "draining" into negative charges. The "amount" of electric field decreases at the same rate as it spreads (since area of a surface increases by the square of its scale). This means that if no matter how we expand our Gaussian surface, if we don't cross ...


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's more the other way around, I would say. Gauss's law, together with the fact that we live in a world with 3 spatial dimensions, requires that the force between charges falls off as 1/r^2. But there are perfectly consistent analogues of electrostatics in worlds with 2 or more spatial dimensions, which each have their own ``Coulomb's law" -- with a ...


4

1) The infinitely long wire has an infinite charge $Q=\lambda \int_{-\infty}^{\infty} \! dz = \infty$, and EM has an infinite range, so one shouldn't be surprised to learn that the result $$\phi(r)~=~ \frac{\lambda}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{\sqrt{z^2+r^2}} ~=~ \frac{\lambda}{4 \pi \epsilon_0} \left[ {\rm arsinh} ...


4

Let us for simplicity consider $n$ point charges $q_1$, $\ldots$, $q_n$, at positions $\vec{r}_1$, $\ldots$, $\vec{r}_n$, in the electrostatic limit, with vacuum permittivity $\epsilon_0$. Now let us sketch one possible strategy to prove Gauss' law from Coulomb's law: Deduce from Coulomb's law that the electric field at position $\vec{r}$ is $$\tag{1} ...


4

Notice that the factors of $r^2$ cancel, and $\hat r\cdot\hat r = 1$ so the integral expression you wrote down reduces to $$ \frac{q}{4\pi\epsilon_0}\int \sin\theta \,d\theta \,d\phi $$ The bounds of integration are $0<\theta<\pi$ and $0<\phi<2\pi$ so we really need to compute $$ \int_0^\pi \sin\theta\,d\theta\int_0^{2\pi}d\phi = 2(2\pi) = ...


4

Contrary to what queueoverflow says, you do not actually need to perform any integration here; a pretty cool symmetry argument will give you the answer. Let the cube we are considering in the problem have side length $\ell$. The trick is to consider putting the charge at the center of an imaginary cube of side length $2\ell$. The flux through the surface ...


4

You are correct: there is no free charge so $\vec{D}=0$ which means $$ \vec{E}=-\frac{1}{\epsilon_0}\vec{P}=-\frac{k}{\epsilon_0r}\hat{r} $$ But this is for $R_1\leq r\leq R_2$. Inside the shell, $r<R_1$, there are no enclosed charges, so $\vec{E}=0$ there. Outside the shell, there is also no charge. Recall that the total charge for dielectrics can be ...


4

You can use $q/\epsilon_0$ to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing. Flux for any closed surface is $\Phi = \oint \vec{E} \cdot d\vec{A}$. There are two ways to calculate this quantity: The hard way, which means evaluating $\vec{E}$ on every part of the surface, and integrating. ...


4

There is indeed a connection. The holomorphy is easily seen in the electrostatic potential. In a charge free (two-dimensional) region, the electrostatic potential solves Laplace's equation and hence is a harmonic function. The real and imaginary parts of a holomorphic function are harmonic functions and thus the electrostatic potential can be identified ...


3

There is a simple reason why we can consider variations on the whole $A$ rather than the quotient $C=A/G$ and the reason is following: all configurations that are $G$-equivalent have the same value of the action $S$. That's what we mean by the statement that the theory has the symmetry $G$. So the variation of the action $S$ in the directions that are ...


3

The magnetic force propagates at the speed of light. If a magnetic dipole were suddenly created the magnetic field lines would be established simultaneously but not in their equilibrium positions. Instead the field lines would bend and oscillate in a similar manar as in an oscillating magnetic dipole. Since magnetic field lines are always loops, even when ...


3

The basic idea, at least as it is taught to the freshmen is as following: In conductor you have plenty of electrons that are practically "free". You are interested in electrostatic situation, that is the situation in which all electrons are by definition practically still. This is possible only if electric field within conductor is zero. If that would ...


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 ...



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