# Tag Info

5

Gauss law says that the total flux going through a closed surface is equal to the charge inside the surface. You can think of flux as the number of field lines going in (or out) through the surface. In your example there is no charge inside the sphere so the total flux through the surface of the sphere is zero. On one end (the left hand side) the field ...

4

The electroscope can be considered a capacitor with capacity $C$, so it will carry a charge $Q = UC$ if we apply a voltage $U$. This means that the needle and the support strut will carry $Q$ and the case will carry the opposite charge. The equal sign charge carried by the needle and the strut repels and equilibrates with the gravity (and constraint ...

4

The charges that accumulate on the plates of a capacitor are not provided by the material of the plates themselves but by the source that is charging them, so there is in principle no limit to the amount of charge that they can hold, if your source is strong enough. The maximum-charge limits on actual physical capacitors are dictated by the dielectric ...

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Let me clarify you what divergence actually is. Consider an arbitrary finite volume $V$, whose surface is $S$, in a vector-field $\bf h$.Then total flux emerging from $S$ is given by $$\text{Net flux from the surface} = \int_S \mathbf{h}\cdot d{\bf a}.$$ Divide $V$ into two parts: $$V = v_1 +v _2$$. Now the flux out of volume $v$ is given by : $\int ... 3 The flat ends of a cylinder are perpendicular to its cylindrical surface. The electrical field is perpendicular to any cylindrical surface centered on the line charge, and so is parallel to the ends of any such cylinder.$\vec E$is the electric field, and I presume$\vec n$is the surface normal of the ends of the Gaussian surface. Since the two vectors ... 3 When we focus on classical mechanics and only take charged particles (mass/ actual electric charge) there is only one difference between the trajectories of the particle in an electrical/ gravitational field: in the electric fields particles can have positive/ negatice charge thus move towards/ away of the source (or to put it that way: in the electric field ... 3 If you brought two identical and perfectly conducting balls together and there were and odd number of excess electrons then one electron can sit at the point of contact while the remaining even number can have half be on each conductor. So each conductor has an equal number of charges. If you then symmetrically pull the spheres away from each other then ... 2 Take the distance between the charges as$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

1) Why the potential at the surface? This approach is probably used because part (a) of the problem gives you an explicit expression for $V_S$ and so the expression for $V(r)$ is self-contained without having to consider what happens inside the shell. Basically the solution makes use of the fact that $$V_S = \int_0^{r_2}{E dr} = \int_0^{r}{E dr} + ... 2 You have more flux per unit area going into the right side, but the area on the right side is smaller. These two balance out so that the total flux is the same going in as going out. The part of the sphere which has electric flux going in, traced in red, is less than half the area of the sphere. Incidentally, flux per unit area is just the electric ... 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 ... 2 Direction of area is in the direction of the normal drawn outwards. Direction of the area and the plane of area are at right angles. 2 Just to expand a little bit on Emilio's answer, and to address the specific point you made in your final paragraph about "finite charges", I decided to calculate what fraction of electrons would be "missing" from the positive plate of a capacitor when you reach breakdown. For a vacuum dielectric, the Schwinger limit is about 1.3\cdot 10^{18}\rm{V/m} - ... 1 In this case, the battery is said to be "floating". Its potential with respect to earth can be suprisingly high or low. Small buildups of static electricity on the battery can easily charge it to hundreds or thousands of volts with respect to earth. The voltage difference across the battery's terminals is still 1.5\,\text{V}, but the voltage of the ... 1 It is true that given an electric field, then you can define uniquely the charge density that created it, by Gauss' law, as you have done. But the converse is not true: given a charge density you cannot define uniquely the electric field that it will create since you have to solve a differential equation (again Gauss' law) to do that and each differential ... 1 Your diagram is a representation of three dimensions on a two dimensional screen... so talking about "up" and "down" is pretty confusing. Think about it this way: if you look from the side, the coaxial cable appears to be two concentric circles right? From this perspective, the set up has ROTATIONAL symmetry(if you rotate the system around the axis, you ... 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 ... 1 I've only ever used the Ewald sum, I've never implemented it myself. However, you mention that you're not converging as \kappa increases nor are you converging to the correct value. It would seem that regardless of the problem, if your implementation is correct it should converge at some point. If you do reach convergence wrt \kappa; as to the point ... 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} - ...

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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.

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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 ...

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$$\rho(\mathbf{r})=\lambda(x)\delta(y)\delta(z)$$ 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 ...

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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. ...

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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 ...

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The first sentence started with an if. When you start with an if and end with a problem a solution you should consider is that your if never happens. So if the EMF around a zero resistance loop is zero then we don't expect the total magnetic flux through to change. Is that reasonable? Yes. Since it is a zero resistance loop, it can generate any current it ...

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It seems that the answer should actually be exact. The proof goes as follows. Let's define charge distribution $C_1$ the distribution the system (the whole system, not just one side) would have, if boundary conditions are satisfied and 1st side of the cube has potential $V_0$. Let's define similarly $C_2$ ... $C_6$. Now these distributions have an obvious ...

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Let's say we hook up a capacitor to a battery (and maybe toss a resistor in there). The battery will pump a charge difference between the plates, which creates a potential difference between the plates. When the potential difference reaches the potential difference of the battery, current that takes and adds charge to the plates stops flowing, as current ...

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Positively charged means lack of electrons. Remember that protons are stationary only free electrons can move. So if you have excess electrons its negative and if no of protons= no of electrons its neutral. So what happens in above case is that one sphere has lack of electrons and the other is neutral. By joining them by a conductor some electrons from ...

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