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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" ...

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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 ...

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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 ...

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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 ${\bf r}_0$ is \tag{1} E^i({\bf r}_0)~=~- \int_{[0,\infty[\times [0,\pi]\times[0,2\pi]} ...

2

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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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 ...

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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 ...

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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.

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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 $3 \;\text{C ... 1 Quite simply, the dipole moment of a charged system depends on the coordinate origin. There is nothing particularly surprising or unphysical about this, and there are plenty of other quantities (such as orbital angular momentum) with that property. The dipole moment of a charged system is not a quantity that is defined for the system itself; it is only ... 1 Assuming that the rings share a central axis A: You'll have a double integral, with the outer integral being a sum of the component of force parallel to the axis A on each tiny chunk dQ of the first ring. The inner integral finds that force component on dQ by adding up the component parallel to A of the force between dQ and each tiny chunk dq of the second ... 1 The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ... 1 Any non uniformity in charge accumulation is supposed to be distributed evenly since the plane conducts. Not true. The charge distribution will be such that there is no component of electric field parallel to the surface of the conductor - because that is what would generate a force on the charges, and cause them to be redistributed. Instead, what ... 1 Much of chapter I, section 9 in "Foundations of potential theory, Oliver Dimon Kellogg, Berlin: Verlag von Julius Springer, 1929", parts of which appear in the answers by Mathaholic, Procyon and Qmechanic, is devoted to answer this question. Let$v$be a small region of arbitrary shape, containing$P$(defined by$\vec{r}$) in its interior. We consider the ... 1 No, there will be a net force because the charged object will induce a charge redistribution in the metal. See electrical induction. 1 In most cases (probably always), no. Suppose the charged object has a positive charge. Then it attracts the electrons in the metal close to itself, creating a positive charge on the region of the metal away from the charged object. The force of attraction on the electrons of the metal is more than the force of repulsion on the positive region of metal, ... 1 why would increasing voltage, while keeping charge constant, have any effect on the ability of a body to store charge. (1) Capacitors don't store charge, they store electrical energy. For a capacitor, it is understood that one plate has charge$Q$while the other plate has charge$-Q$so there is no net electric charge stored. (2) If you increase ... 1 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 ... 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}\$. ...

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