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1

For a Gaussian surface between the two plates, the total flux through the surface is zero. For the particular surface you give, all of the electric field lines crossing one of the circular faces cross the other face but in an opposite sense. This is because the outward normal vector for one face of the cylinder is opposite in direction to the outward ...


5

Your equation is wrong. The flux is not $2EA$. Rather, is $-EA + EA = 0$. The reason is because Gauss's law involves the dot product of $E$ and $dA$. The direction of $dA$ is the way out of the cylinder. When the electric field goes through the bottom of the cylinder, the electric field is in the opposite direction of the area vector, so the electric flux is ...


2

Because Gauss's law applies for both moving and stationary charges, while Coloumb's law applies only for stationary charges, Gauss's law can be considered more fundemental. This is why Gauss's law is one of the four Maxwell equations. The derivation of Gauss's law from Coloumb's law only works for stationary charges; for moving charges the derivation is ...


0

From the Feynman Lectures on Physics (I would have made this a comment but I don't have enough points) From our derivation you see that Gauss' law follows from the fact that the exponent in Coulomb's law is exactly two. A $1/r^3$ field, or any $1/r^n$ field with $n≠2$, would not give Gauss' law. So Gauss' law is just an expression, in a different ...


0

A charge cannot come out of nowhere. The various possibilities are: The charge must have existed beforehand in which case most regions of space would have had an electric field already, or the charge was created as part of pair of opposite charges, in which case the electric field lines would curve back to the opposite charge, and any flux entering the ...


0

You can resolve the E vectors coming out of the yellow surface into a component parallel to the surface and a component perpendicular to the surface: $E = E_{parallel} + E_{perpendicular}$. As I say to my students a lot of the time, draw the triangle! The triangle here has a hypotenuse of one of the E vectors (in red), one leg of the triangle is the ...


1

Your problem is the assumption "both the charges contribute flux to their respective Gaussian surfaces only at the common surface." When you take your surface out to infinity you are also increasing its area, so even though the electric field goes to zero, the integral of the electric field over your surface will be non-zero. In fact, the total flux ...


2

You have two distinct errors. One is claiming that because the electric field goes to zero at infinity so does the flux. The flux is the integral of the electric field over the surface. The electric field goes down as $\frac 1{r^2}$, but the area goes up as $r^2$, s the flux constant. Ask Gauss about this. The second is to claim that because (from the ...


1

Along with Timaeus and hft, I wonder what you would consider to be acceptable assumptions. Usually Coulomb's law is taken to be less universal than Gauss' law, so one is usually more interested to derive Coulomb's law as a special case of Gauss' law for static charges rather than the other way around. If you want to reverse that and start from Coulomb's ...


0

If you had Maxwell's other equations $$\vec{\nabla}\cdot\vec{B}=0,$$ $$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t},$$ $$\vec{\nabla}\times\vec{B}=\mu_0\left(\vec{J}+\epsilon_0\frac{\partial\vec{E}}{\partial t}\right),$$ then for smooth enough fields (enough that partials in different directions commute) you can take the divergence of ...


1

Now if I choose an imaginary enclosed surface in that part of space would the electric flux through that surface sum out to zero owing to absence of charge there? Yes. I meant to ask can we mathematically see whether it would sum up to zero thereby confirming Gauss's law No, we can't confirm Gauss's law this way. To understand these ...


1

I have posted an answer before, but, thinking about it for a while, deleted it in shame. I still have no answer and even more questions, but OP and readers will probably enjoy this article on the subject: Notes on Gauss’s law applied for time varying electric field in vacuum, published on arXiv, 26 Jan 2015. It contains formal derivation of flux of electric ...


1

You're not particularly spot on with your definition of electric flux. Most fundamentally, the electric flux $\Phi$ through a given surface $S$ is defined to be $$ \Phi=\int_S \mathbf E\cdot\mathrm d \mathbf S. $$ If you introduce a well-defined model in terms of field lines, then this does end up describing the number of field lines that cross $S$, to ...


0

As you say: "If there is any stray charge outside of a certain conductor, couldn't there be some on an internal surface of a huge, hollow conductor?" Certainly! I think your confusion arises from a misunderstanding of what "outside" means in this context. Outside does not mean "the furthest surface from the center of mass of an object." It means "the ...


0

I think the correct answer for your question is that the electric charge outside of a closed Gaussian surface does not enter inside because the surface is enclosed. Take a look at the Faraday's cage and you can also take into account why we are not affected when driving in a car and get struck by a lightening.


1

The electric field in the problem has no $z$ component, so it quite simple to calculate the flux through a cylinder with axis parallel to the $ z $ axis; then you choose a cylinder that contains the sphere you are interested in. Let $\Sigma$ be the surface of the cylinder, $ V $ its volume, $\Sigma '$ and $ V' $ the surface and volume of the sphere; by the ...


1

For proving that the Coulomb E satisfies the Gauss law in differential form, simply take the divergence of Coulombian E. You will need the Dirac delta and some of its properties. For deriving the divergence of E field, do just the same as above... You can also derive the Gauss law in integral form from the differential form by applying the divergence ...


0

In your decompositon you have $$ \psi = 1/r^3$$ and $$ \vec{E} = \vec{r} \, .$$ Then, as you wrote, $$ \vec{\nabla} \cdot \left(\vec{r} \frac{1}{r^3}\right) = \vec{\nabla}\psi \cdot \vec{r} + \psi \vec{\nabla}\cdot \vec{r} \, .$$ Insert $$ \vec{\nabla} \psi = -\frac{3}{r^5} \vec{r}$$ and $$\vec{\nabla} \cdot \vec{r} = 3$$ and you are done. Note ...


1

Consider a single charge $q$ (out of a given system of charges) at a point $\textbf x$ in free space. The charge $q$ will feel electrostatic forces due to all the other charges. To analyze how $q$ will behave, we have to consider the force that acts on it, which is $$ \textbf F = q \textbf E,$$ where $\textbf E$ is the electrostatic field generated by all ...


2

If you don't specify the boundary, there will be many solutions (including the unsymmetric ones) The problem of this question is that solving Maxwell's differential equations necessarily involves specifying the boundary conditions which usually can be chosen the obvious ones, however here such a boundary simply does not exist. Usually, it is ...


1

I believe that this is a wrongly posed problem. The proof goes as follows: due to the isotropy and homogeneity of the configuration, the field has to be zero everywhere, because there is no preferred direction in your problem. The field has a direction, s.t. if it weren't null one should have a preferred direction in your problem. Let's now apply your ...


3

Let's be really clear about the exact kind of mistake you are making. Consider the numbers $1,-1,1,-1,1,-1, \dots$. If you wanted to sum them you could argue that $1-1+1-1+1-1+\dots=(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0$. Your friend could argue that $$1-1+1-1+1-1+\dots=1+(-1+1)+(-1+1)++\dots=1+0+0+\dots=1.$$ The problem is that an infinite series isn't ...


17

Gauss's law is always fine. It is one of the tenets of electromagnetism, as one of Maxwell's equations, and as far as we can tell they always agree with experiment. The problem you've uncovered is simply that "a uniform charge density of infinite extent" is not actually physically possible, and it turns out that (i) it is not possible to express it as the ...


-1

Try turning the problem on it's head. Start with a sphere of charge density $\rho$, with empty space outside. The electric field is as you have calculated it. Now, fill up the rest of space with charges. Each time you place a new charge $q$, which causes an electric field $\vec{E}_1$ at some point on the surface of the original sphere, add another charge $q$ ...


0

First, the context is electrostatics so we assume the net force on any charged particle is zero; if the net force is not zero, the charged particle will accelerate and the situation would not be static. Inside a conductor, by definition, charge is free to move; if there is an electric field inside, charge will be accelerated. Thus, we must conclude that, ...


0

I will begin from the end. In the book where you see $\Phi = \oint\vec{E} \cdot d\vec{A}$ the meaning is a closed surface. The same meaning has the symbol in Internet and the circle/ellipse on the integrals means closed, i.e. a closed surface. From the element of integration you understand that you have to do with a surface. About the presence of the ...



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