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Assuming that the rings share a central axis A: You'll have a double integral, with the outer integral being a sum of the component of force parallel to the axis A on each tiny chunk dQ of the first ring. The inner integral finds that force component on dQ by adding up the component parallel to A of the force between dQ and each tiny chunk dq of the second ...

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One way to think of it is to imagine a point source pushing out water. If you put some imaginary surface around the source, the total amount of water passing through that surface is the same as the amount of water being pushed out by the source. It doesn't matter what the shape of this surface is, the total amount of water passing through it will be the same ...

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If I'm understanding your question correctly, it's because we design the Gaussian surface that way. Here's what I mean by that: Gauss's law works for any arbitrary closed surface, regardless of its shape. The total flux through the surface must be proportional to the enclosed charge. For most surfaces you could invent (and you really are free to invent ...

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Quite simply, the dipole moment of a charged system depends on the coordinate origin. There is nothing particularly surprising or unphysical about this, and there are plenty of other quantities (such as orbital angular momentum) with that property. The dipole moment of a charged system is not a quantity that is defined for the system itself; it is only ...

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Forget anything about a capacitor and just consider the resistance of the conducting liquid. Think of the liquid as made up of thin $dr$ concentric shells of radius $r$ and find the resistance of a shell in terms of the resistivity, radius and thickness. Then do the integration to find the resistance of all the liquid.

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For instance, why don't measure the ability to store something by the volume it takes so why not charge per unit volume. There is nothing wrong with you defining a parameter which is the "charge per unit volume" but after defining it then what are you going to do with it? So here you have a capacitor and its charge per unit volume is $3 \;\text{C ... 2 You have misrepresented the citation in the book. The 5th edition page 757 discusses experiments with a hollow sphere and a solid sphere. The experiments verify that the exponent is 2 within experimental error. 0 Alright, one issue I noticed is the incorrect Gauss' Law expression. It is actually: $$\oint {\vec E}\cdot \mathrm{d}{\vec A} = \frac{q}{\epsilon_0}$$ So only the perpendicular component of the vectors through each face surface matters. Since the electric field vectors for all the voltages you've given across each surface are acting upwards, only the ... -1 This mostly comes under the theories of matter and therefore essentially chemistry i.e the types of bonds. The typical bonds are metallic bonds, ionic bonds, covalent bonds, hydrogen bonds, as well as Van de Wall forces and the like. I advise looking into these. Electromagnetism is responsible for the amalgamation of elements into compounds of any phase, ... 0 The quantity of electrons traveling across this interesting capacitor is necessarily hindered by the formula you used. Although there would be this consideration for electron emission, the potential through the dielectric would be chiefly concerned with the polarization and I would therefore consider that. Polarization is the net shift in dipole moments of ... 2 I understand that capacitance is the ability of a body to store an electrical charge and the formula is$C = {Q \over V}$Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ... 2 We Use$C=Q/V$because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ... 0 Determine the surface element of the cube, 2x2y2z or 8xyz, then use the integral version of Gauss's Law. 1 The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ... 20 You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ... 2 A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ... 4 Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$on one side,$-Q$on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ... 1 why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge$Q$while the other plate has charge$-Q$so there is no net electric charge stored. (2) If you increase ... 1 In most cases (probably always), no. Suppose the charged object has a positive charge. Then it attracts the electrons in the metal close to itself, creating a positive charge on the region of the metal away from the charged object. The force of attraction on the electrons of the metal is more than the force of repulsion on the positive region of metal, ... 1 No, there will be a net force because the charged object will induce a charge redistribution in the metal. See electrical induction. 1 Much of chapter I, section 9 in "Foundations of potential theory, Oliver Dimon Kellogg, Berlin: Verlag von Julius Springer, 1929", parts of which appear in the answers by Mathaholic, Procyon and Qmechanic, is devoted to answer this question. Let$v$be a small region of arbitrary shape, containing$P$(defined by$\vec{r}$) in its interior. We consider the ... 0 First off, what is meant by the potential of the conductors? The "absolute" potential is of no importance in a working capacitor. In fact, it's practically impossible to determine. The potential difference between the plates (or between a plate and some ground/earth plane) is the important factor in the capacitance relationship,$C=qV$. Now ... 0 In all the cases, there will be same answer, the charges will equally spread among the outermost surface of the objects. It is proved by Faraday's butterfly-net experiment. And even if it be conducting or non-conducting, the charge will be equally spread over the surface and here we get charge density. And in most cases, the charge density is more important ... 0 Consider the two spheres to be conducting, with the inner sphere of radius$a$and the outer of radius$b$. If the outer sphere carries a charge$Q$, then the inner will have a charge$-\dfrac{a}{b}Q$. In fact, the charge on the outer one will create a potential$V=\dfrac{kQ}{b}$on its surface and inside. Therefore, to keep its zero potential, the inner ... 0 If the shell is connected to the earth, it will have no charge. It is placed inside another shell. If the unearthed shell is charged, then there will be that much charge on it. But there will be no charge on the inside shell whether it is earthed or not. It is because there is no electric field inside a charged shell. The charges can reside only on the ... 0 It looks like I can answer at least one question. The charge imbalance in the carpet will dissipate into the air via ionized moisture, as described in this question: How does an object regains its neutrality after being charged by rubbing? 0 Consider a metal sphere of unit positive charge. Now , place a point charge at centre of sphere. What do you observe? You find that electric field inside sphere is not zero. Why? Because , electric field due to charge on metal is cancelled and not by charge inside metal sphere. Returning to your question Similarly, Electric field due to ... 1 Any non uniformity in charge accumulation is supposed to be distributed evenly since the plane conducts. Not true. The charge distribution will be such that there is no component of electric field parallel to the surface of the conductor - because that is what would generate a force on the charges, and cause them to be redistributed. Instead, what ... 0 I assume you are talking about linear materials with dielectric constant greater than$1$. Say you have a free charge distribution$\rho$in vacuum, which produces an E field. Now you introduce linear material. Then the free charges will be "weakened" because they will by partially screened by charges of the dipoles sticking on them. However, the opposite ... 0 (For completely filled capacitors) Q = CV So, C = Q/V So, C is charge stored per unit Potential Difference applied. Now, V = Ed ,where d is distance between plates.$E = \dfrac{V}{d}$Case 1) When you apply a constant V of 1V to capacitor E across capacitor is$ \dfrac{1V}{d}$which is constant independent of capacitance of capacitor or dielectric b/w ... 0 a) Capacitance C increases b) Charge Q remains unchanged c) if charge Q is constant while C increases, that means voltage V decreases (C=Q/V). d) U decreases ( U=Q^2/2C) 2 We put (four pi times) the permittivity$4\pi\varepsilon=1$equal to one from now on (in the SI unit system) for simplicity. Let${\bf r}_0\in \mathbb{R}^3$be a fixed position. The electric field in the$i$'th Cartesian direction at${\bf r}_0is \tag{1} E^i({\bf r}_0)~=~- \int_{[0,\infty[\times [0,\pi]\times[0,2\pi]} ... 1 I agree that the volume form in spherical coordinates cancels the 1/r^2 divergence from Coulomb's law, but I think there's a more physical way to interpret this mathematical fact: remember that strictly speaking, for a continuous charge distribution with no delta functions in the density, \rho({\bf x}) is not the amount of charge at point {\bf x}. ... 1 I assume that A is neutral to begin with. Then inside of A cannot be any charge by Gauss law. This means that the inside of A must neutral. Then you can take the volume between B and A. The boundary of that are the metal spheres A and B. On the whole boundary surface, there cannot be any electric field as that surface lies inside the metal. If there was any ... 0 Potential energy is associated with a system, not a particle. One must calculate potential energy for pairs of interacting particles, not individual particles. F_{ext} is a force external to the system. It can do external work and raise the energy of the system. F_R is work internal to the system, and does not increase the energy of the system. But ... 2 The electric field at a point outside the volume charge distribution is well defined, certainly. What about the electric field at a point inside the charge distribution? Let \vec{r'} denote a point that lies in the charge distribution and \vec{r} denote a point where the electric field is to be determined. The problem is the determination of the electric ... 0 For a discrete system of N charges, the potential energy associated with their configuration is given by \begin{align}U&= \frac12\,\sum_{j=1}^N \, q_j\sum_{k\ne j}\,\frac1{4\pi\epsilon_0}\,\cdot \frac{q_k}{r_{jk}}\\ &= \frac12\,\sum_{j=1}^N \, q_j\,\varphi(\mathbf r_j)\tag 1 \end{align} where \varphi is the scalar potential due to all charges ... 0 You have{U = \frac{1}{2}\frac{Q^2}{\frac{\epsilon_{0}A}{d}}}{\frac{\epsilon_{0}A}{d}}=C\equiv\textrm{ the capacitance of the capacitor.}$$Hence$${U=\frac{1}{2}\frac{Q^2}{C}}$$Now \displaystyle{Q=CV}, where V is the p.d. between the capacitor plates. So,$${U=\frac{1}{2}CV^2}$$or$${U=\frac{1}{2}\frac{\epsilon_{0}A}{d}V^2}$$... 0 If it wouldn't be its tangential component would exert a force on the charges and they would move. The condition would then not be static. After some movement and redistribution of charges (when there would be no force on charges) the condition will again become static thus making the field only normal to the surface 0 I agree with gatso, but I think it might be explained more simply by noting that the volume element can be written as dV=4πr^2.dr which cancels out the r^2 in the denominator. I also agree that the problem is similar to integrating (1/√x)dx from x=0, because here the ratio dx/√x = 2√x.d(√x)/√x = 2d(√x). 5 I actually agree with Ruslan's comment. You cannot say that the integral blows up when \textbf{r} = \textbf{r}', with \textbf{r}' spanning the integration domain where \rho \neq 0. The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function. To see this you ... 3 I would say yes, the electric field is well defined in a volume charge distribution. You raise some interesting points, that I think ultimately comes down to: Is the electric field consistent and well defined mathematically? What is the physical interpretation of the mathematics? Is a volume charge distribution physical and realistic? To answer the first ... 0 (Ah, a very old question which is not yet answered... and I don't find a duplicate answer, either, though the problem is very nice ... the comment is point you there, and that the user "centralcharge" has edited the question is amazingly fitting, too :) but it's not yet elaborated) Yes, you can use the method of images. Normally you are right, it requires ... 3 As stated by Lemon, electric flux through a volume enclosed by a closed surface is zero when the volume contains no net charge. Electric flux through a closed surface \rm S is$$\Phi= \int_{\mathrm S} \,\mathbf E\cdot \mathbf n\,\mathrm d^2 \mathbf r\;.Now, according to Divergence Theorem, \begin{align}\int_{\mathrm S} \,\mathbf E\cdot \mathbf ... -1 Use the concept of solid angle and equate the two fluxes then find any angle you want to find... 1 For a charge distribution \rho({\bf r'}), the electric field at {\bf r} is{\bf E}({\bf r})=k\iiint\frac{\rho({\bf r'})}{|{\bf r}-{\bf r'}|^2}\hat{\bf R}d^3{\bf r'}$where${\bf R}={\bf r}-{\bf r'}$. I think the OP's claim is at positions where$\rho \ne 0$, the above integral is infinite because the integrand blows up at${\bf R}={\bf 0}$. I think ... 2 The electric potential of a point in space is defined by the mechanical energy that it takes to get a (positive) unit charge to that point. We usually define the potential of an infinitely distant point as zero, and then the movement of our test charge is from infinity to the point for which we want to measure the potential. In case of a capacitor the ... 0 When a conductor is placed in an electric field (uniform for ease) a force exerted by this field acts on the charges in the sphere which pulls hem apart according to the applied field. The charges separate and creates a field until equilibrium is reached. A charge q on the sphere experiences a force$F=qE$. The electrostatic force between two charges acts ... 0 First of all, electrostatic field is conservative in nature. So this force can be written as the gradient of a scalar potential V$E=-∇V$The d in your equation is not right. The gradient indicates the rate of change of V with respect to the three coordinates along the three directions. Since the electric field is zero, we can write$∇V=0$which ... 0 "Potential at a point is sort of energy you used to move a point particle."It is work done to move a unit positive charge in a field. Consider a shell with radius r and charge q. Potential at infinity is taken as 0 by convention. When you move a charge from infinity to surface of shell , you did work$W = \frac{kq}{r}\$ in present of electric field E. ...

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