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## Hot answers tagged electrostatics

12

The maximum charge a capacitor stores depends on the voltage $V_0$ you've used to charge it according to the formula: $$Q_0=CV_0$$ However, a real capacitor will only work for voltages up to the breakdown voltage of the dielectric medium in the capacitor. So in reality, for every capacitor there is a maximum possible charge $Q_{max}$ given by: $$... 10 Loosely speaking, as we walk away from a sphere it looks smaller, as we walk away from a cylinder just the radius looks smaller, but not the infinite length, and finally as we walk away from an infinite sheet of charge it never looks any smaller (we can never 'get away' from an infinite sheet). At more mathematical level I would say the best way to see ... 10 Electrical analogies of mechanical elements such as springs, masses, and dash pots provide the answer. The "deep" connection is simply that the differential equations have the same form. In electric circuit theory, the across variable is voltage while the through variable is current. The analogous quantities in mechanics are force and velocity. Note that ... 9 Short answer: this is a textbook example of the limitations of ideal circuit theory. There seems to be a paradox until the underlying premises are examined closely. The fact is that, if we assume ideal capacitors and ideal superconductors, i.e., ideal short circuits, there appears to be unexplained missing energy. What's not being considered is the ... 7 Yes. The delta function always has the same dimensions as the inverse of its argument. You can read this from its definition, your first equation. So in one dimension \delta (x) has dimensions of inverse of length, in three spatial dimensions \delta^{3}(\vec x) or simply \delta(\vec x) has dimension of inverse of volume, and in n spatial dimension ... 7 In an attempt to be brief: The big thing to remember is that the flux is also proportional to the area (technically, the surface integral of the field over the area). Crudely speaking, the side of the enclosed surface with exiting field lines are further away from the external charge than the side with "entering" field lines, and the surface area increases ... 6 There does seem to be a lot of mythology around about the "grape in a microwave" experiment. I have never see any publications on the subject in a respectable journal, however from chatting to other scientists there seems to be a consensus about what happens. It's all rather boring really. The grape is the right size (about a quarter wavelength) and shape ... 6 This problem with N point charges on a sphere is a famous problem in electrostatics known as the Thomson problem. For large N, it is in general an open problem still under active research. References: Wikipedia.org Mathworld.wolfram.com Mathpages.com 6 Yes, the field is infinite, but it is only log divergent near the plate, so that it is hard to see the divergence numerically. You can see this easily by solving the problem of a uniformly charged infinite plate, which is a 2d problem. Here the charges are uniform along the negative real axis, where the 2d space is imagined to be the complex plane. This ... 6 Four possibilities come to mind, in decreasing order of feasibility: Is barefoot an option? I'm willing to bet it will significantly mitigate the buildup of charge. The two of you only experience a shock on contact because you are at different electric potentials. If you can't keep him at your potential, why not try to join his? Before helping him down, ... 6 Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates. First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the z-direction. I will follow the notation of the Wiki article:$$ x=a\cosh\mu\cos\nu\cos\phi\\ y=a\cosh\mu\cos\nu\sin\phi\\ z=a\sinh\mu\sin\nu $$where ... 6 I usually find it easier to use model multipoles that are surface charges on a sphere, rather than point charges on some polyhedron's vertices. These charge densities are given in general by$$\sigma_{lm}(\theta,\phi)=N\cos(m\phi)P_l^m(\cos(\theta)).$$Thus a monopole is constant, a dipole has \sigma=\cos(\theta)=z/r, a quadrupole has the form ... 5 Electrons will flow against the electric field lines because their charge is negative, and the electric field thus exerts a force \mathbf{F}=q\mathbf{E} on them which is in the opposite direction. Thus electric field lines inside the wire go from the positive to the negative terminal and the electron flow goes from the negative to the positive terminal. ... 5 Conservation of angular momentum does not predict that the disk stays motionless, because the field in this case has angular momentum. The charges produce an electric field, and the magnetic field is not parallel to it, so there is a Poynting vector going around in circles, and the field angular momentum is just converted to mechanical angular momentum when ... 5 Moving mass does generate gravitation different from stationary mass. This is the ''gravitomagnetic'' effect predicted by Lens and Thirring in the 20's and measured by Gravity Probe B: http://en.wikipedia.org/wiki/Gravitoelectromagnetism It is related to the ''frame dragging'' effect that you hear about with respect to spinning black holes. There, there ... 5 General Relativity is a mathematical model that relates the curvature of spacetime to an object called the stress-energy tensor. In many cases the stress-energy tensor is dominated by mass and you can simply consider the curvature as being related to the mass. However this isn't always true as I'll mention below. Anyhow, we can put any numbers we want into ... 5 The device you describe is called a Jacob's Ladder. You are correct that it is high voltage between the rods that produces the initial spark at the bottom of the ladder where the gap between the rods is the narrowest. Then the ionized air heats up, becoming less dense, so it rises. The current path rises as well because once a breakdown of the air has ... 5 As the two charged bodies attract, they have unlike charges. So, Assuming your two charged bodies as conductors and charged equally, the system may be considered as a Capacitor. If you place a dielectric like glass of some Relative permittivity \epsilon_r (3.7 to 10) which fills the empty space between the bodies, then the capacitance would be ... 5 Electrons just don't like each other, a point captured by the phrase that "like charges repel." So, imagine a gymnasium full of students pretending to be electrons, staying as far away from others as possible. Anyone near the center of the crowd will feel badly pressed and will try to work there way towards the edge of the gym, where at least one side will ... 5 Consider a charged conductor made out of two spheres of radii R_1 and R_2, connected with a conducting wire. Assume that R_1<R_2, and that the spheres are far apart so that effects of electrostatic interactions between the spheres can be neglected. Then, the surface charge density, the quantity that describes how crowded the charges are, is higher ... 5 The nature (and glory) of the dirac delta function is that the volume integral$$ \int_{\Delta V} dV' \delta ( \boldsymbol{r-r'} ) = \left\{ \begin{array}{cc} 1 & \text{if } \Delta V \text{ contains } \boldsymbol{r}\\ 0 & \text{if } \Delta V \text{ does not contain } \boldsymbol{r} \end{array} \right. $$Using this function, you can write the ... 5 Crazy Buddy's answer and related comments have made the point that you could indeed use a capacitor to charge a battery, but the amount of energy stored in capacitors is generally less than in batteries so it wouldn't charge the battery very much. However there is a new generation of capacitors called ultracapacitors that are being developed with electric ... 5 If it is in air (or any other substance), there is a limit where the electric field of the object is going to be enough to ionize the surrounding medium, and the resulting current will drain the object of its charge. Similarly, if the object is immersed in vacuum, you will eventually have an electric field sufficient to "polarize the vacuum" by creating ... 5 Since this is a homework-type problem, here are some Hints for the force The electrostatic force d\vec F on a small segment dl of the rod given the field \vec E of the other rod is$$ d\vec F = \lambda\, dl \,\vec E $$Determine the field of one rod, and use the above expression to integrate the force it exerts on the other rod. This is a 2D ... 5 First of all note that k is not dimensionless, it is k = \frac{1}{4 \pi \varepsilon_0}, and \varepsilon_0 has dimensions of \frac{ \text C^2}{ \text {N m}^2}. So you have already \frac{ \text{V C}^2 \text m^2}{ \text {N m m} ^2}. Also, volt can be expanded as  \text V = \frac{ \text {N m}}{ \text C}, so one gets$$ \frac{ \text C^2}{ \text {N ...

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The exact solution is $${\bf E}(R<r, \theta =\pi/2)=\frac{Q}{4 \pi \epsilon_0 }\left(\frac{1}{r^2}\right)\sum_{l=0}^{\infty}\frac{(2l)!}{2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l}\hat{{\bf r}}.$$ Clearly the field inside the conductor (that is, for $r<R$) vanishes. Here $Q$ is the total charge on the disk. The field, for large values of $r$, looks ...

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It's not intuition.It's a problem which can be solved. First we identify the sign of the charges. By seeing the direction of field lines we can see that the sign of charges. Field lines originate from $+ve$ and end at $-ve$ charges. Next by Definition of Flux, The number of field lines cutting per unit surface surface . And Gauss' Law The flux ...

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Important notice: My previous result was a little bit incorrect. I found the factor $1/2$ by comparison with the textbook V.V. Batygin, I.N. Toptygin «Problems in electrodynamics». Let's denote the radius of the inner sphere $S_1$ as $a$, the radius of the outer sphere $S_2$ as $b$ and the displacement as $c$, so that $c\ll a,b$. We choose the origin of the ...

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A battery generates a voltage by a chemical reaction. There is a class of chemical reactions called redox reactions that involve the transport of electrons, and you can use the reaction to drive electrons through an external circuit. This is the basis of a battery. The battery will continue to provide power until all the reagents have been used up and the ...

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