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The fact is not only that is infinite, but that it is a plane. Consider the homogeneous charged point (i.e., a point charge): $E \propto \frac{1}{r^2}.$ In the homogeneous charged (also infinite) rod: $E \propto \frac{1}{r}$. In the picture below, if you consider the effect of the yellow area of the plate right under a probe at a distance $d$, the ...


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You have two questions, and they have different answers. First of all, let's be clear about what Gauss's law is in integral form: $$ \int \vec{E} \cdot \mathrm{d} \vec{A} = \frac{Q_\mathrm{encl}}{\epsilon_0} $$ In words: the total flux integrated over a closed surface is equal to the charge enclosed in that surface, divided by the permittivity of free ...


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Using distributions is a trick that people use whenever dealing with systems that do not have the required smoothness and integrability conditions and is in general only a mathematical technique to nevertheless solve those problems. Maxwell equations, to start with, require both sides to be differentiable (at least a few times) and integrable and when you ...


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The "unique" here is, IMO, not a good, evocative word. A better one would be preferred direction or privileged direction. Another way of looking at this is all directions are equivalent. A further confusing subtlety is that there is also something else that the author is assuming without telling you. There are no privileged directions for a problems with ...


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The object for which you need to find the electric field is a uniformly charged sphere. Uniformly charged means that at every point on the sphere the charge density is same. Suppose someone blind-folded you and then he rotated the sphere in some arbitrary fashion about the origin(assuming your sphere has origin as the center). Then he takes off the blind ...


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I think, in order to generate the field $E$ there should be some charge distribution inside the cube. For example an electromagnetic wave propagates through vacuum, that is, there is no charge necessary to generate it, at least with fitting boundary conditions. But for a static electric field with non-zero divergence, there should be some charge ...


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Consider the three-terminal device that is your stacked capacitor: A ----============================================= (dielectric medium ɛ) =============================================---- B (dielectric medium ɛ) C ----============================================= All three plates have the same area A. It's ...


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There is a much simpler way to find the charge on the conductor. Using the method of images, the image charges are just the negative of the charges above the plane. Therefore, the total charge on the surface of the plane is -Q. The first-term in your dipole field is wrong. The unit vectors have unit value and do not cancel r squared in the denominator.


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Gauss' Law says $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ whereas the total vector area is $\iint_S \mathbf{dA}=\mathbf{0}$ for some closed surface $S$. The total vector area is taking vector sum of all the differential area vectors that are normal to the surface, whereas $\iint_S \mathbf{E} \cdot \mathbf{dA} = Q/\epsilon_0$ is taking the dot ...


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When you find the vector area of a closed surface you are dealing with vectors and the total sum is zero. However, when you use Gauss' law with a constant E what you are integrating is a scalar, and that gives you the surface (not zero).


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According to my knowledge, Gauss Law can be applied to any function where the quantity(the force) is inversely proportional to distance squared.Now, is the force required to be conservative?


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Indeed, we have that: $$\nabla \cdot \vec{g}=-4\pi G\rho$$ Where $\rho$ is the mass density. Integrating both sides, we have: $$\iiint_{\Sigma}(\nabla \cdot \vec{g})\:\mathrm{d}V=-4\pi G\iiint_{\Sigma}\rho(\vec{r})\:\mathrm{d}^{3}\vec{r}$$ But by Gauss' Law (the divergence theorem), we have: $$\oint_{\partial \Sigma}\vec{g}\cdot\mathrm{d}\vec{A}=-4\pi ...



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