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## Hot answers tagged electrostatics

3

The flat ends of a cylinder are perpendicular to its cylindrical surface. The electrical field is perpendicular to any cylindrical surface centered on the line charge, and so is parallel to the ends of any such cylinder. $\vec E$ is the electric field, and I presume $\vec n$ is the surface normal of the ends of the Gaussian surface. Since the two vectors ...

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If you brought two identical and perfectly conducting balls together and there were and odd number of excess electrons then one electron can sit at the point of contact while the remaining even number can have half be on each conductor. So each conductor has an equal number of charges. If you then symmetrically pull the spheres away from each other then ...

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Take the distance between the charges as $D$, and we'll assume that the dielectric constant, $K$, at a distance $x$ between the charges is given by some function $f(x)$. Actually let's assume that $f(x)$ gives $\sqrt{K}$ as this makes the notation clearer. If you take some small element $dx$, then the effective length of this element is $\sqrt{K}\, dx = ... 2 It depends on what exactly you are asking. Suppose we take 64g of copper i.e. one mole of copper. Each copper atom contributes one conduction electron, so our chunk of copper contains$6.023 \times 10^{23}$(Avagadro's number) conduction electrons with a total charge of 96488 coulombs. John's answer involves removing those electrons by a chemical reaction. ... 2 These are all the same equation, because$C = \frac{Q}{V}$. You can convert. As long as you know any two of$Q$,$C$, and$V$, you can use the equation that references those two, or use the formula above to convert to another one. 2 This is one of the places where we can make things perfectly rigorous if we make certain assumptions on the charge density$\rho$(and$D$). I will rigorously show you in the following that$\Vert \mathbf{E}(\mathbf{x}_{0}) \Vert < \infty$for all$\mathbf{x}_{0}$in the interior of$D$, assuming that$\rho \in (L^{1} \cap L^{\infty})(D)$. The only ... 1 An alternative and shorter answer is that the expression you cite for$\bf{E}(\bf{x}_0)$makes use of the Green function$\bf{G}(\bf{x}; \bf{y}) = \bf{G}(\bf{x} - \bf{y})$satisfying the distributional equation $$\nabla_{\bf{x}} \cdot \bf{G}(\bf{x}-\bf{y}) = \delta(\bf{x} - \bf{y})$$ and reading $${\bf G}(\bf{x} - \bf{y}) = \frac{\bf{x} - ... 1 You "guess net charge will get distributed equally between the two objects", but that guess is wrong. If you touch the objects and then separate them again, one might wind up with charge 3 and the other 2, or 4 and 1, or 1000 and -995, or whatever. 1 You have been misled, I think, by the term "unit of charge". This can have two different meanings depending on context. Three, actually. At the microscopic (but not the quantum) level, the unit of charge is the charge on an electron or a proton. If you interpret the question as implying this unit of charge, then your question is quite reasonable, and your ... 1$$\rho(\mathbf{r})=\lambda(x)\delta(y)\delta(z)$$describes a charge density in the form of a (possibly infinite, depends on what your allowed x values are in the system) line in 3D space, where$\lambda(x)$is the linear charge density as a function of x. The delta functions indicates the charge density is concentrated at one point in the yz plane, but ... 1 In QED, this superposition principle is still valid at least to first order perturbation theory. The deeper reason behind this superposition principle from the QED point of view, is, that photons do not interact with each other. They do not carry charges. They do not "see" each other. So they can be safely superposed without having an effect on each other. ... 1 I believe the following picture explains what's missing: The cylinder is "infinite", but the Gaussian surface that is drawn as part of the analysis is in the shape of a finite cylinder with flat ends. And since the electric field is at every point perpendicular to the wire, it is parallel to these flat ends. Parallel to the surface means perpendicular to ... 1 In all conductors, charges reside on the surface. The reason for this is that conductors have free electrons, that is, the electrons are loosely attached to the nucleus of the atoms in the conductors. Refer to the pic below (drawn in MS Paint) When placed in external electric field, the electrons migrate to one side of the conductor and an electric ... 1 The sheet is grounded, so$V(x,y,0)$is null, while boundary conditions at$z \rightarrow -\infty$also impose$V(x,y,z\rightarrow -\infty) \rightarrow 0$. Since there is no charge in the half-space below the sheet, the potential there can only be$V_{below}(x,y,z) = 0$. So$\frac{\partial V_{below}}{\partial n} = 0\$ as the charge density expression ...

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