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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

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 ...


6

As the very formulation of your question makes clear, we know what the actual algebra of local symmetries is. It is the five-dimensional diffeomorphism invariance assuming the $M^4\times S^1$ topology of the five-dimensional spacetime. The term "Kač-Moody generalization of an algebra" is nothing else than an alternative name for this algebra, especially for ...


3

Group actions in classical field theory. Let a classical theory of fields $\Phi:M\to V$ be given, where $M$ is a ``base" manifold and $V$ is some target space, which, for the sake of simplicity and pedagogical clarity, we take to be a vector space. Let $\mathscr F$ denote the set of admissible field configurations, namely the set of admissible functions ...


2

All of those charges on the other side of the sphere "conspire" to exactly cancel the field of the nearby charges on the surface. You're analysis is correct. This result can also be shown by integration of Coulomb's Law, but it's not an easy calculation.


1

About the supposed paradox: $u$ and $\bar d$ have the same isospin quantum numbers, but not all the other properties. If you restrict your problem to only study the isospin space, you will not see that they have different charge and other different quantum numbers. About the charge: I don't know where your equation comes from, but it seems close to the ...


1

By applying Gauss' Law one gets (the surface integral over the sphere with $r>R$): $$\oint_s \vec{E}(r, \theta, \phi) \cdot \hat{n}(r, \theta, \phi) \, ds = \oint_s E(r, \theta, \phi)\, ds = \iiint E(r, \theta, \phi)\, r^2 \sin\theta \, d\theta \, d\phi = E4{\pi}r^2$$ The surface integral depends only on $r$ and is equal to the area of the sphere. $E$ ...


1

You can say this by spherical symmetry argument. All points with given $R$ on a sphere are equivalent. Now if $E$ depends on $\theta$, then it is different in different directions. So if someone says that $E$ is more at an angle of $30^0$, then your reply would be: "why not $45^0$ or $70^0$. What is so special about $30^0$." And that is it. There is ...


1

Emmy Noether proved both the theorem and its converse. Look for the book "The Noether Theorems" for a precise and discussed formulation of her statements, as well as a translation of the original paper. It seems there is a link to the pdf in the princeton math website (I don't know about copyright issues, however).


1

Pick a point above the plane. From a point in the plane directly under the point above, draw a circle of some radius. Consider the contribution of the charge elements along the circle to the electric field at the point above the plane. Since the charge density is uniform, the horizontal components of the electric field from charge elements on opposite ...


1

An answer connected to Gauss law (I hope everything is correct, since it's long ago for me ... so no warranty): An infinite plane of uniform charge for example in the z-plane has the charge distribution: $\rho=q\,\delta(z)$ Thus, the electrostatic potential should be $\Phi=\frac{q\,|z|}{2\pi}$. Hence, the electric vectorfield is: ...


1

The best way to make intuitive, is to draw columb force vecrtor from any chosen point of the sphere, to the point you measuring the field in, then project this vector on three chosen coordinates (it's better to choose spherical one) , then chose the oposite point (relative to sphere radios) and do the same projection, and u will see how magically, all ...



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