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Your reasoning is correct, it's just a lot harder to do with the surfaces you've chosen. Draw a small, elemental ring at some arbitrary height above the charge. A line from any point on the ring to the charge subtends the polar angle $\theta$ with the z-axis and is a radial distance $r$ from it. (i.e. $r$ and $\theta$ are the usual spherical polar ...

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You probably did some wrong calculation, because your reasoning works. Take a circle of radius $r$ a distance $a$ above the charge. The increase $d\Phi$ of the flux by increasing the radius by $dr$ is given by $$d\Phi=\frac{q}{4\pi\epsilon_0}\frac{2\pi r \cos \theta dr}{a^2+r^2}=\frac{q}{2\epsilon_0}\frac{ar dr}{2(a^2+r^2)^{3/2}},$$ where $\theta$ is the ...

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The Gauss Law indicates that the field lines $\vec{E}$ should be normal to the Gaussian Surface taken $dA$. Thus we take the dot-product to take the normal component of field $\vec{E}$ with the area. $\vec{E}\cdot \hat{n}dA = EdAcos\theta$ The reason to take Gaussian surface as a sphere, with the point charge being its center, is because 1) the field ...

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This question is using spherical coordinates, and in spherical coordinates the divergence of a vector (like $\nabla\cdot \textbf{E}$) is: $\nabla\cdot \textbf{A}=\frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2 A_r \right) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta} \left( A_\theta\sin\theta \right) + \frac{1}{r\sin\theta}\frac{\partial ... 0 The electric field given in the problem only depends on the radius$r$. Therefore, the divergence is most conveniently computed in spherical coordinates. The factors of$r^2$and$1/r^2$come from the transformation of$\nabla$to spherical coordinates, see here. For a function$f(r)\hat{r}\$ the divergence is $$\nabla\cdot ... 0 Look at the expression for the divergence in the spherical coordinates. 2 Here's a simple way of looking at it: If you are close to an infinite plane, you may be feeling stronger attraction by every individual part of it, but "more" of those parts are pulling you at a significant angle. This way, a lot of the attraction is canceling out. As it happens (this is anything but coincidence though), these two opposite effects exactly ... 0 You are confusing work on a closed loop, with an integral on a closed surface. What is true is that for eletrostatics, we have$$\oint_C\mathbf{E}\cdot d\mathbf{l}=0,$$where C is a closed curve, which is Ampère's law. What Gauss law states is that the electric flux over a closed surface is equal to the charge enclosed by it. 3 Given that \rho is the charge density, the integral,$$\frac{1}{\epsilon_0}\iiint_{V} \rho\, dV = \frac{Q}{\epsilon_0}$$Now, Gauss' law states that,$$\iint_{\partial V} E \, dS = \frac{Q}{\epsilon_0}$$Hence, we arrive at your 'global form' by simply equating:$$\iint_{\partial V} E \, dS = \frac{1}{\epsilon_0}\iiint_{V} \rho\, dV$$By the ... 3 \oint E\cdot dS = \frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV  if \frac{1}{\epsilon}\int\limits_V \rho dV= \int \limits_V \nabla \cdot E \space dV then \frac{\rho}{\epsilon} = \nabla \cdot E if \rho=0 then \frac{\rho}{\epsilon} = 0 = \nabla \cdot E Is this what your looking for? \rho would be zero say, ... 2 That's probably for charged solid sphere, not a cylinder. In any case, setting the potential at infinity as zero, we have for r>R:$$V(r)-V(\infty)=-\int_\infty^rE(r')dr'\implies V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$$For r<R, we got:$$V(r)-V(\infty)=-\int_\infty^rE(r')dr'=-\int_\infty^RE(r')dr'-\int_R^rE(r')dr' \implies \\ ...

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