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## Hot answers tagged electrostatics

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This is a good example of a procedure that happens in many areas of physics. In general, physical laws - and particularly conservation laws - tend to be most naturally phrased in integral form, or even in mixed integro-differential form. For an example of the latter, consider the integral form of Faraday's law: $$\oint_{\partial S}\mathbf{E}\cdot\text ... 5 The electric field assigns a single vector quantity to each point in space (specifically, the direction in which a positive test charge would accelerate if it popped into existence at that point, assuming it didn't perturb the setup creating the field in the first place). I believe that the difficulty of this question arises from an ambiguity in the problem ... 4 Suppose you have two conductors kept at voltages V_1 and V_2 and they have charges Q_1 and Q_2 respectively. Then, one has the relation$$ Q_1 = C_{11}\ V_1 + C_{12}\ V_2\quad \textrm{and}\quad Q_2 = C_{12}\ V_1 + C_{22} \ V_2 \ , $$which defines a (symmetric) Capacitance matrix that is determined by the geometries of the two conductors. This ... 4 If the rods were really far apart then the positive charge would be equally distributed throughout each rod. If you push the rods together then the new equilibrium involves fewer charges bunched around the closer points and more of them at the far ends; the energy you exert is the energy it takes to move these charges around. Even if we didn't consider the ... 4 Here's one way to think about it (though it isn't mathematically rigorous). From very far away the dipole would appear to have zero charge and thus there wouldn't be an electric field at all. However, you also know that the electric field falls off as 1/r, so from very far away you'd expect the electric field to be small. The additional charge ... 4 Faraday's cage is known to block static and non-static electric fields. The mechanism of blocking depends on whether the electric field is static or non-static (EM field). I suppose you question is about how the cage works in non-electrostatic case. In EM case (time changing field), two scenarios could happen. The first is electric discharge where the ... 4 Well, you have think about the definition of capacitance, as dmckee pointed out in his comment. For two conductors both charged with charge Q and at a potential difference V, capacitance is$$ C = \frac{Q}{V} $$So capacitance is a proportionality constant between charge on two conductor and the potential difference. Now, if you consider two parallel ... 3 A simple example is that of a sphere. One way to find its capacitance is to take the limit of a nested sphere capacitor with radii a,b:$$C = \lim_{b\to\infty}\frac{4\pi\epsilon_0}{\frac{1}{a}-\frac{1}{b}} = 4\pi a\epsilon_0\text{.}$$A van de Graaff generator is a commonly discussed in physics classes, and involves this type of setup. For a ... 3 The electrostatic potential energy of a system of point charges is defined as the work required to be done to bring the charges constituting the system to their respective locations from infinity. Suppose we have a configuration of point charges. If the potential of the energy of the system is negative, this means work is positive. Consider two point ... 3 If you break the electric field lines you've drawn into components parallel to the surface and components perpendicular to the surface, you'll find a net field pointing to the left. That means positive charges would feel a force to the left, and since this is a conductor, they're free to move, and so they move to the left. The only way for the charges to ... 3 The plates form a capacitor from the beginning. (More precisely a coupled capacitor with three terminals because you have also to regard the neutral ground as one of the terminals.) The capacity coefficient between the plates is low at the beginning since the plates have a large distance. Therefore, you need a high voltage to put the charge on them. In the ... 2 Dfg, you have not drawn an equilibrium electric field and the charges will move. We can decompose the field you drew into two parts -- the perpendicular part, and the horizontal part. This horizontal part pushes the charges to the left. You are confused about the field lines not intersecting the surface. In this case they do. Outside of equilibrium, fields ... 2 Rule of thumb for working it out: If you imagine letting a charge go, the direction it tends to move is toward lower potential energy. The opposite direction is toward higher potential energy. This is independent of the choice of where the zero of energy is. 2 As Mostafa says, it is macroscopically at equilibrium, not necessarily microscopically. There may be one misunderstanding you have, which is about "surface". I will talk about it later. In my opinion, equilibrium should be understood as no electron moving. It is easily to show that the electric field in conductor is zero. If the electric field is non-zero, ... 2 Systems of plates are not typically considered capacitors unless they are globally neutral. Nevertheless, capacitance is a geometric property that is to do with the system more than the actual voltages and charges you apply to it, so that your question still makes sense: the capacitance is the same as it would be with symmetric charges. More specifically, ... 2 I personally think the text is misleading. It's blindly applying Gauss' law while not considering its subtleties. Here's a more cause-and-effect way to look at it. After this, we'll get to Gauss' law. Let's take a look at the positively charged plate. Yes, the surface charge density on one side doubles. But the surface charge density on the other side goes ... 2 You simplified incorrectly, the result is zero. I assume you computed it using floating point arithmetic, which explains why the discrepancy is on the order of machine epsilon. Try using symbolic manipulation in Mathematica: Sum[Cos[2 \[Pi] n/13], {n, 13}] // FullSimplify Out: 0 1 You just made some math mistakes. You made a mistake when you did Q = h\int_A kr. You got Q = h\pi k r^2, but you should have gotten Q = \frac{2}{3} h\pi k r^3. Notice how this second expression has units of charge while the first one doesn't. Another mistake you make is that you say \frac{1}{r} \frac{\partial rE(r)}{\partial r} = \frac{1}{r} ... 1 The first one is electrostatic potential energy, the second on is electric field. You can tell they are supposed to represent different physical quantities because they have different units. I am pretty sure that the way it is presented in your textbook, the second equation for the electric field is to be seen as justified by experiment and you will derive ... 1 This can be solved in a single line of math:$$\left\langle\rho_1,\phi_2\right\rangle=\left\langle\rho_1,\nabla^{-2}\rho_2\right\rangle=\left\langle\nabla^{-2}\rho_1,\rho_2\right\rangle=\left\langle\phi_1,\rho_2\right\rangle$$where the self-adjointness property of \nabla^{2k} has been used, and \left\langle\right\rangle represents the integration inner ... 1 The field itself carries energy. This is, in fact, a vital point because it can be shown that, if the momentum and energy carried by the fields isn't accounted for, electromagnetism would blatantly violate Newton's third law (and this doesn't have anything to do with special relativity per se). 1 Equilibrium in the sense of this question means there are no net forces on the objects that make up the system: the charges contained in the conductor. Note that we need a model of an ideal conductor here. A neutral ideal conductor is thought of as containing equal large amounts of unbound, infinitely small (not electrons) positive and negative charges. ... 1 There are two contributions to the electric field in a dielectric: The field generated by the 'free' charges, i.e the ones on the capacitor plates. Call it E_0 E_0 polarizes the dielectric, which in turn adds to the total electric field. Call that polarization P. The total electric field is$$E=E_0-\epsilon_0^{-1}P (The factor of ...

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Initially when you attach the capacitor to the battery, said battery will act to create an electric field within the wire. On the side of the negative terminal this field will point perpendicular to the cross section of the wire toward the terminal of the battery (electric field points toward negative charge). On the side of the positive terminal the field ...

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If the electric and magnetic fields are steady, $\oint \mathbf{E}\cdot\mathrm{d}\mathbf{l} = 0$ is a law of nature. Even if you change the material across the wire, the electric field should still respect $\oint \mathbf{E}\cdot\mathrm{d}\mathbf{l} = 0$ and therefore must be uniform. In fact, this is one of the boundary conditions at dielectric interfaces, if ...

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He used what is called a torsion balance. His experimental method is outlined very nicely in this video. After Coulomb published the result of his work, however, it was debated as to whether his experiment really did provide enough evidence to support his claim that the force between two point charges really did follow the equation we now call Coulomb's ...

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With work you also have to take care in specifying what is doing the work. If it is the work required to move a particle or the work done by the electric field, these have different signs. If the work done by the electric field on a point charge is positive it means it is moving in the direction of the electric force acting on the point charge, therefore, $W ... 1 Have to be a bit careful with potential energies, as the 0 point of potential can be arbitrarily chosen. Only changes in potential are well defined without a choice of 0 point. That said, it is often convenient to choose the 0 point at$\infty$, and this is the typical choice when talking about assembling point charges. With this choice, the potential ... 1 If there is an electric field, then there will be a force on a charged particle since$\vec{F} = q \vec{E}\$. If you put a single charged particle in perfect vacuum, then if you add another charged particle then both particles must either move away from or towards one another, depending on whether the charges of the two particles have the same or opposite ...

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