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This is presumably because the shaded-blue region is a conductor and the electric field in there must be zero; all of the field lines of the inner charge must therefore terminate on positive charges at the surface of the conductor. The field in a conductor must be 0 because if it weren't it would generate a large current (for a perfect conductor, an ...


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For external points, cylinder behaves like a line of charge. So you just have to put the distance in the formula of electric field of a point charge as (radius + distance from the cylinders surface)


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No. The charge enclosed by a surface is really just the number of protons inside minus the number of electrons inside all multiplied by the charge of the proton (which is a constant). If the electric field has a large density in one region then it has less in another so that total flux over a closed surface is still the net enclosed by the whole surface is ...


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A conductor can have a zero net charge. One way to do this is to have an inner surface with a net charge of $-Q$ and an outer surface with a net charge of $+Q.$ So for instance you can have a surface charge density of $-Q/(4\pi R^2)$ on the surface $r=R$ and you can have a surface charge density of $+Q/(4\pi (2 R)^2)$ on the surface $r=2R$ and this has zero ...


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First, we will compute $\text{div}\,F$. The partial derivatives are given by$${{\partial F}\over{\partial x_i}} = q {\partial\over{\partial x_i}}\left({{x_i}\over{r^3}}\right) = q\left({1\over{r^3}} - {{3r^2 {{x_i}\over{r}} x_i}\over{r^6}}\right) = 0.$$Thus, $\text{div}\,F = 0$ away from the origin. Consider now a ball $B$ of radius $r$ centered at the ...


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Considering the charge in an enclosed surface is always 0 I'm not exactly sure what you mean by this but do understand that the Gaussian surface 'encloses' a volume of space within which the enclosed charge resides. For example, consider an isolated point charge $q$. Due the spherical symmetry, the appropriate Gaussian surface is a sphere of radius $r ...


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(1) If there is net electric flux through the closed surface then some electric field lines either originate or terminate within the volume enclosed. (2) If no electric field lines originate or terminate within the volume enclosed, all of the field lines entering the volume exit the volume and there is thus zero net electric flux through the closed surface. ...


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Imagine it this way, keep a charge $q$ near the cube and let the field lines enter the cube. Now we know that field is a function of $r$ and varies as $1/r^2$ for a point charge which means that electric field should not be zero unless $r$ is infinite and if it is not zero, the field lines entering will be equal to field lines exiting although the intensity ...


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Your reasoning is correct, but you are not accounting that the flux is the multiplication of $\bf E$ and $d \bf S$, and $d\bf S$ $ = r^2 d\bf\Omega$ increases with $r^2$. So this compensates the decrease of $\bf E$ with $r^2$.


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Well for the three sides touching the particle, the vector $\bf E$ coming from the particle, is always contained in the surface. This means that there is no component in the direction normal to any of the surfaces. So this means $\bf E$$d \bf S$ and therefore the flux through them, $\int \bf{E}$ $d \bf S$ will result zero.


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1) One eighth of the charge is contained in the cube because you can place eight cubes such that they have a common vertex, the one your charge is at. If your charge was on the middle of an edge, it would be shared by four cubes. If your charge was in the middle of one of the faces, it would be shared by two cubes and so on. 2) The three adjacent faces have ...


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Suppose you have a charge in a plane. The electric field from the charge is spherically symmetric, and that means every field line from the charge that intersects the plane has an equal and opposite field line intersecting the plane. So when we integrate $\mathbf{E}\cdot\text{d}\mathbf{A}$ the two field lines will cancel out and the net flux will be zero. ...


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My question is how the three sides don't contribute any flux whereas we can see that a small fraction amount of flux can pass through the three sides. Not really. You see, the electrostatic field $\vec{E}$ of the charge is always radially outwards. If the charge is situated at the exact corner of the cube, then the field is exactly coplanar with the ...


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Hint: You can always construct imaginary cubes (7 more, in this case.) such that your charge is completely enclosed inside your multi-cube system. You can find the flux through this system via Gauss' Law. The total flux through each side (of the new configuration) would be equal via symmetry.


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First question: I do not have Griffiths book but I hope this is what you mean. You can use Gauss law on a surface that encloses the surface charge, you just can't use it on the surface itself, because the electric field is infinite there. Mathematically you can state: $$\iint \mathrm{d}S \, \, \vec{n} \cdot \vec{D}= \sum_i \iiint_{V_i} \mathrm{d}V \, \, ...


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The main flaw in your argument is that you consider them to be a single rectangular conductor. In reality, metals form layers of oxides on their surface, which is why when I place two metal cups together they don't just 'join up' and become one piece of metal. To actually join the metals, you would have to weld them together, in which case they are no longer ...


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We know that inside a conductor the electric field $E$ is always equal to $0$. Hence there cannot be any net charge inside the conductor. As like charges repel, the charges, being equal to their neighbors tend to go as far away from each other as possible. The only configuration where both the above conditions are met, is when the net charge resides on the ...



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