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1

there will be no tangential component of E^ in the case of the charged surface in electrostatics. If so then this would result in a force on the free electrons which would cause a drift hence a current which is not desired in electrostatics.


2

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 ...


2

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 $$d\Phi=\frac{q}{4\pi\epsilon_0}\frac{2\pi r \cos \theta dr}{a^2+r^2}=\frac{q}{2\epsilon_0}\frac{ar dr}{(a^2+r^2)^{3/2}},$$ where $\theta$ is the ...


1

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 ...


0

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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