# Tag Info

27

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 ...

21

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 ...

19

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 ...

16

Field lines draw all of their validity from Gauss's law for the electrostatic field, $$\nabla\cdot \mathbf{E}=\frac1{\epsilon_0}\rho,\ \text{or equivalently}\ \oint_{\partial\Omega}\mathbf{E}\cdot\text d\mathbf{S}=\frac1{\epsilon_0}Q_\Omega,$$ where $Q_\Omega=\int_\Omega\rho\,\text d\mathbf{r}$ is the electric charge in a volume $\Omega$ whose surface is ... 14 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 d\... 12 One of the reason's I like to cite for this is empirical observation. For years we have observed that the planets in the solar system move in closed orbits (roughly atleast). Mind you, I haven't asked for the orbits to be circular or elliptic, just closed. Now, invoking Bertrand's theorem with regard to the Kepler problem of the motion of a body in a central ... 12 One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be ... 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 ... 11 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 ... 11 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 ... 10 This become a lot clearer if you consider the integral forms of Maxwell's equations. We start with Gauss' Law $$\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}$$ If we integrate this over some volume V and apply Gauss' Divergence Theorem we find that the left hand side gives \begin{align} \int_V\mathrm{d}^3\vec{x}\;\nabla\cdot\... 9 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 ... 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 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. ... 7 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 ... 7 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 ... 7 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 ... 7 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 ... 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 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 ... 7 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 ... 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. ... 7 Actually, your expression for the potential \Phi(r) is incorrect. The expression \Phi(r) = -\frac{GM(r)}{r} is only valid outside the sphere. As an explicit demonstration of its invalidity, note that\underset{r\rightarrow0}{\text{lim}}\,\Phi(r)=\underset{r\rightarrow0}{\text{lim}}\,\left[-\frac{G}{r}\int_0^r4\pi r'^2\rho(r')\,dr'\right]=0$$assuming ... 7 It doesn't hold for arbitrary shapes. The reason it works for spheres is that when you have a spherical charge distribution and a concentric spherical Gaussian surface, the whole system is invariant under rotations around the center of the spheres. If the electric field were different at different points on the Gaussian sphere, you could rotate the whole ... 7 A lot of things decrease in intensity as 1/r^2, such as light intensity, gravity, charge forces, etc. This is because the same force needs to act over a larger spherical area. The further away, the larger the sphere. And you should know that the surface area of a sphere is SA=4\pi r^2. Since the area varies as r^2, dividing the magnitude of the ... 7 We have Newton's law in the form$$ F = \frac{Gm_1m_2}{r^2}$$which is the same as the field equation for the potential$$ \nabla^2 \phi = 4\pi G \rho $$where \rho is the mass density. The 4\pi here does indeed come from the solid angle of an entire sphere, but by redefining G we could put it in Newton's law instead. This is connected to GR by ... 6 The equation$$\displaystyle\int_{V}\frac{\rho}{\epsilon_0}d\tau=\int_{V}(\nabla\cdot E)~d\tau$is true for all region$V$in space the integration is performed over. That is why it follows that the integrands are equal. Your counterexample is invalid, because the integrals are equal only when the domain of integration is of the form$[-a,a]\$.

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: ...

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