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## New answers tagged gauss-law

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By "the designation of 1 and 2 is interchangeable" it means the flux passing through the upper surface equals the flux passing through the lower surface. In more details, because your system has reflection symmetry about the $xy$ plane and $E$ field is a vector, $\textbf{E}(x,y,-z)$ should be the mirror image of $\textbf{E}(x,y,z)$. Hence $$\textbf{E}_z(x,... 4 For point sources of a field or energy source, such as a charged particle, a gravitational body (which acts like a point source), or a loudspeaker on top of a tall column, the geometry of the problem controls how energy and fields distribute themselves in space. At all points that are an equal distance from a point source, the energy or field strength is ... 0 Gauss law in 2D would have to be:$$\oint \mathbf{E} \cdot \mathbf{\hat{n}} dl = 2 \pi q$$because you are reducing your surface in 3D to a line in 2D, and keep the idea of measure of the boundary and its orthogonal direction or normal. To get the expression of the field you have to make use of the fact that the electric field is isotropic. In other words,... -1 Note-: My grammar isn't that good. But with the help of diagrams, I hope everyone will understand. Lets consider any closed surface in space with the charge (or origin) anywhere inside it. Lets also consider an infinitesimal solid angle from the origin. Let dS be the infinitesimal area by which an infinitesimal solid angle is subtended.(Since there are ... 2 Here is another attempt at an answer, from a slightly different angle. It is less sophisticated and may add insight. You say that the capacitors are isolated before they are connected. Literally, that is only possible if they are in different universes. In the real world, they are connected through an infinitesimal capacitance - the capacitance between the ... 0 If the shell is conductive in this case, one thing for sure is that the net charge on the interior surface has the opposite sign and equal amount of charge as the charge placed inside of the cavity. If there are no net charge on the conductor before the charge inside of the cavity was placed, the charge on the conductor equal to the charge placed inside the ... 2 Your statement is correct. The charge distribution is such that the hollow cavity of the conductor has a equal amount of negative charge induced on its inner part. This distribution is such that field due the cavity (including the charge inside the cavity) cancels out everywhere outside the cavity. So looking it the other way around external sources do not ... 1 For simplicity I treat this as a two-dimensional problem. The diagram below shows a dipole of length l and a Gaussian surface of radius R. At point A the dot product \vec E_{R1}\cdot \vec R is positive and at point B the dot product \vec E_{R2}\cdot \vec R is negative where \vec E_{R1} and \vec E_{R2} are the net electric fields at points A... 2 Fundamental particles like the electron are well-known to have magnetic dipole moments related to their spins. Moreover, the standard model also predicts that they should have small electric dipole moments (EDMs) as well. However, the EDMs predicted in the standard model are very small; they will probably be too small to observe directly for quite some ... 0 I think it is not the fastest solution to try and use the divergence theorem, try to apply \nabla two times. Take in your notation$$\nabla^2\Phi = (-G) \Delta \iiint_D\frac{\sigma(\xi)}{||x-\xi||}\ d\xi\ = (-G) \iiint_D \sigma(\xi) \Delta \frac{1}{||x-\xi||}\ d\xi\ $$If I remember my field theory lectures correctly$$\Delta \frac{1}{||x-\xi||} = 4\pi \...

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The Electric field outside a point of the shell=KQ/r^2 assuming Q=Charge on the shell and r=distance of the point from the Centre of the Sphere.This can be easily derived if you draw another Gaussian sphere of radius r enclosing the given Sphere.By Gauss Law, Closed integral of E.dA=Charge enclosed/Epsilon. And You can easily find out the Electric Field. ...

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First and foremost, I don't think you could ever prove Gauss's Law wrong. If it says the field(and hence force) is zero, it is indeed zero, and accordingly its consequences. Now if you want the physical situation that causes the force experienced inside to be zero, it is crudely represented in the diagram. Consider the major bowl and the minor bowl(i like ...

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If you know Gauss's Law, apply it. Within the boundary, charge enclosed is zero. So, closed integral of E.dS is zero(i.e. flux). Now, we know that the field and area vectors are parallel, and area vector is non-zero. What is indeed zero is the magnitude of field vector, and there you go. Field inside the spherical shell is zero at any point. On drawing a ...

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Yes, and relativistic gravity neatly explains Gauss's law. The simple way to think of gravity resulting from spacetime curvature is: a 3 dimensional volume of space containing mass M expands over time at a volume acceleration of 4πGM. That's essentially a relativistic re-expression of Gauss's Law where instead of describing a gravitational field it ...

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Suppose the surface charge densities on the bottom plate is $\sigma$ and on the top plate $-\sigma$, then the electric field due to the bottom plate is $\frac{\sigma}{2 \epsilon_0}{\bf n}$ and that due to the top plate $-\frac{\sigma}{2 \epsilon_0}{\bf n}$, where ${\bf n}$ is a unit vector pointing from the bottom plate to the top plate. This gives the total ...

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