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2

Indeed, the $\vec{E}$ field in a parallel plate is independent of distance from the plate. This works because of the assumption $d \ll$ length of plate (thus, we can ignore side effects of the plate). And as Bort pointed out, it is the Voltage $V$ that scales linearly with respect to distance from the plate, while $\vec{E}$ will remain constant.


3

You need to watch what you mean by the ambiguous term "derive", which can mean either "was derived historically" (i.e. was motivated by or is a derivative of, in the non-mathematical sense) or "is derived logically/mathematically". Historically, I think you are correct that $\boldsymbol{\nabla}\cdot ...


0

Well I dont know if we can prove it but there is a much more elegant way of formulating EM which may be helpful here. As you may know there are two potentials on EM: the scalar potential $\phi$ and the vector potential $\vec{A}$, from which $\vec{E}(t,x)$ and $\vec{B}(t,x)$ are derived. From this two objects and following symmetry considerations you can ...


3

Maxwell derived his equations from 1) charge conservation law; 2) Coulomb's law; 3) Bio--Savart--Laplace law; 4) Faraday's law of induction. The equation $\boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0}$ was indeed derived from Coulomb's law and in its differential form is written using Gauss--Ostrogradskiy theorem. ...


3

No we cannot prove it; Maxwell postulated that it would hold dynamically because it made the most sense for it to do so as he pondered the displacement current problem. As you likely know, Maxwell pondered the inconsistency between Ampère's law for magnetostatics and the charge continuity equation. Ampère's law for magnetostatics reads $\nabla\times ...


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The answer to this question completely depends upon how the charges are distributed.Saying that Gauss law holds here is correct but not in all cases. Suppose that the positive charges are to be randomly placed. Such a distribution would have slight non-uniformity, which would develop further with time because positive repels positive. So, a net flux will ...


5

The problem here is that you've failed to specify a boundary condition. Consider an electrostatics problem where you're given a charge distribution $\rho(\mathbf{r})$ and asked to find the electric field $\mathbf{E}(\mathbf{r})$. The electric field is the solution to the set of differential equations $\nabla \times \mathbf{E} = 0$ and $\nabla \cdot ...


1

This is a good question because you are about to learn something. As outside charge create electric fields going in, it then also contributes fields going out on the opposite side of the surface. So they cancel themselves out. Where as the charge inside the surface only has electric fields going out.


0

It means that he might as well be in outer space. He will float weightless in the air. If he jumps toward the center of the earth he will gradually slow down and stop due to air friction, and will then be trapped, unable to move in any direction. Unless he drank a lot of fluids before he jumped.


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If the shell is perfectly spherically symmetric, and the charge is perfectly evenly distributed on it, then the total field due to every singel charge is zero at every single point inside the sphere. This also happens with gravity, there is no gravitational force inside a uniformly distributed shell of mass due to the shell of mass. While each piece of ...


2

The statement means that the net electric field at any given point inside the sphere adds up to zero due to all the varying contributions by the charges on the surface. They exactly cancel out, and hence for any point inside the sphere, the value of electric field is exactly zero.


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This question is answered in the Wikipedia article on the shell theorem. The gist is that it is not the total. But going beyond that, I think your question actually reflects a misunderstanding of what "electric field" means. The electric field is something which has a value at every point in space. If you try to calculate a total, e.g. by integrating the ...


0

Why is it said that Gauss's Law is mainly applicable for symmetric surfaces/bodies? Why not for asymmetric surfaces? Nobody says that. Gauß law holds whenever its hypotheses hold. In particular it has nothing to do with the shape of the bodies at hand, rather it is a mathematical theorem relating the flow of a vector field through a surface to the ...


1

As an example, let us suppose that you want to use the Gauss law to evaluate the electric field generated by a body charged in an uniform way. The gauss law tell you that the flux over an arbitrary closed surface around your body is proportional to the total charge: $$\int_{\partial V} \vec{E}\cdot d\vec{S}=\frac{Q}{\epsilon_0} $$ but this is an ...


1

As your teacher says, it holds for every surface, but a look at the law itself, should clear out why some form of symmetrie is desirable: $$ \iint_S \vec{E} .\mathrm{d}\vec{A}=\iint_S E . \cos\theta . \mathrm{d}A = \frac{Q}{\epsilon_0} $$ Here, $E$ and $\theta$ are position-dependent, so to calculate the integral, you need to take care of a position ...


1

The answer to your question involves the fact that one does not usually know a priori the electric field $\textbf{E}$ (or, for that matter, its direction) of a charge distribution $\rho$. Gauss's law, in integral form, relates the flux of the electric field through some closed surface $S$ to the charge enclosed within the volume bounded by $S$. Precisely, ...


2

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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