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22

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 ... 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} ... 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 ... 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 ... 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 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 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$$assuming ... 7 If I understood your question correctly, then you want a simple experiment to demonstrate that magnetic monopoles cannot exist. The simplest way to explain this to a high schooler would be to actually break a small piece of magnet, and then make the student realize that the poles of the magnet haven't been 'split'; instead, both the pieces contain two poles. ... 6 I can give you an intuitive view from a physicist. Charges are the sources and sinks for the electrical field. Consider the extreme case where the volume enclosed by the surface is empty space, so no charges. Then any field line that enters the volume must exit the volume somewhere else. Thus, the integral of the field over the entire surface is 0. If ... 6 I would say yes ! Actually some theories explaining quantum gravity use also this reasoning: gravity is a very weak interaction at a quantum level because it "leaks" into other dimensions, not observable at our scale, but that are present at this scale. The mathematical tools are different, but if you just think about gauss's law you can imagine one ... 6 Gauss' law is applicable for a finite wire. But, it's useless in this case. In the infinite example, you assume some things due to symmetry, namely: It's pretty obvious why these things can be assumed--moving up and down the wire should not change \vec E, so we take it constant. Also, there should be no direction bias, so \vec E has no component ... 6 In the equations as you've written them, the constant of proportionality is an outward-pointing vector for the electric field and an inward-pointing vector for the gravitational field. Or in other words, if you take the radial component only: it's a positive constant for the electric force and a negative constant for the gravitational force. The details: ... 6 I) Right, the differential form of Gauss's law$$\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).$$Note in particular, that it is technically wrong to claim (as ... 5 If you followed the arguments carefully and checked what is demonstrably right and what is not, you would agree that what the argument actually does is to prove that a uniform electric charge density cannot have a uniform electric field. Your original task was to solve Maxwell's equations (well, Gauss's law), so if you find out that the equations aren't ... 5 The field line picture known from school might be helpful with that: The surface area of the surrounding sphere (and not it's volume) determines the density of the lines sourced by a point charge, corresponding to the field strength. 5 \oint_{\partial V} g\cdot dS=-4\pi GM_{enclosed} where integral is surface integral over a closed surface. 5 You are correct: there is no free charge so \vec{D}=0 which means$$ \vec{E}=-\frac{1}{\epsilon_0}\vec{P}=-\frac{k}{\epsilon_0r}\hat{r} $$But this is for R_1\leq r\leq R_2. Inside the shell, r<R_1, there are no enclosed charges, so \vec{E}=0 there. Outside the shell, there is also no charge. Recall that the total charge for dielectrics can be ... 5 You tell us that one surface of the box is at B, but you're a little vague on where the opposite face is. You do say that your surface is "between the two sheets", so I think you may mean that the surface is entirely contained in the space between the two sheets. The box does not intersect any charged surface. With that, and a uniform electric field in ... 5 Gauss' law and Coulomb's law are equivalent - meaning that they are one and the same thing. Either one of them can be derived from the other. The rigorous derivations can be found in any of the electrodynamics textbooks, for eg., Jackson. For eg., consider a point charge q. As per Coulomb's law, the electric field produced by it is given by$$\vec{E} = ...

5

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 ...

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

Maxwell's equations in curved spacetime are written in the form $$\begin{split}\nabla_a F^{ab} &= - 4\pi J^b,\\ \nabla_{[a} F_{bc]} &= 0,\end{split}$$ with $F$ the Faraday two-form, $J^a$ the current four-vector, $\nabla$ the covariant derivative and $[]$ denotes antisymmetrization of the indices. In terms of exterior calculus they become: $$... 4 You can use q/\epsilon_0 to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing. Flux for any closed surface is \Phi = \oint \vec{E} \cdot d\vec{A}. There are two ways to calculate this quantity: The hard way, which means evaluating \vec{E} on every part of the surface, and integrating. ... 4 Notice that the factors of r^2 cancel, and \hat r\cdot\hat r = 1 so the integral expression you wrote down reduces to$$ \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 ...

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