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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 ...


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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 ...


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Look at the expression for the divergence in the spherical coordinates.


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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 ...


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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.


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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 ...


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$\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, ...


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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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