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Distribution of charge within the region where the field is located, is obviously uniquely defined, because it's just $$\rho=\epsilon_0\nabla \vec{E}$$ However, if you cut out a region of space, and want to predict the contents of this region based only on the field outside, you can't do it in a unique way. The reason is that you have too many degrees of ...


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Static electricity is a subfield of classical electromagnetism, the theory of electrically charged particles as well as electric and magnetic fields. Photons only enter the picture in the quantum version of the theory, Quantum Electrodyanimcs (QED). EM waves are not mediated by photons, but EM waves and (real) photons are one and the same thing. Static ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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If I assume electric field lines form closed loop that would mean electric field has non zero curl. So I cant write electric field to be gradient of some scalar function. That would imply work done by the electric field will depend on the path. we know thats not the case really. Another way to see this: Closed electric field lines would mean number of ...


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If there are no charges inside the cylinder, then the potential obeys Laplace's Equation: $$\nabla^2V = 0$$ In cylindrical coordinates, that's $$\left(\frac{\partial^2}{\partial r^2} + \frac 1 r \frac{\partial}{\partial r} + \frac{1}{r^2} \frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}\right) V = 0$$ Based on your notation, it seems ...


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I don't see any problems up until maybe the very very last step (where you lost some constants at the very least and I can't tell what you were trying to do or why you think there is a problem). So you were at: $$\int_{-\infty}^{\infty}\frac{\sigma}{2\pi\epsilon_0 (x^2+z_0^2)} x dx.$$ I'm not sure if you then tried a u-substitution or just found an ...



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