# Tag Info

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How does one calculate the capacitance of two bodies with different charges? To be clear, capacitance doesn't depend on the amount of charge; the capacitance is determined by the geometry of the bodies. If you have two conductors, there are actually three capacitances to consider, the self-capacitance of each and mutual capacitance of the two ...

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I was wondering why, in the derivation, they set charge to Q... but if the definition is amount of charge being able to be store $Q$ is the charge separated, not stored. A charged capacitor is not electrically charged any more than a charged battery is. Rather, a charged capacitor has stored energy. To charge a parallel plate capacitor, electric ...

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By charge on capacitor, we (generally) mean the magnitude of charge on one of the plates.

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You have to be careful of your Gaussian surface. If you include both charges (from both plates) then your Gaussian surface is outside of the parallel plate capacitor and the electric field is indeed zero because there is zero net charge when encapsulating the parallel plates with your Gaussian surface. However, if you place one part side of your Gaussian ...

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Since the plate is a conductor the surface charge is spread across both sides of plate (see fig 23.16 a or b). Hence this gives that the electric field in between the two plates is actually $$E=\frac{\sigma}{2\epsilon_{0}}$$ This comes from Gauss' Law $\oint \vec{E}\cdot d\vec{A}=\frac{Q_{enclosed}}{\epsilon_{0}}$ Now, the electric field for two ...

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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 ...

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A Short answer is here, Firstly the capacitor gets charged.Though it takes infinite time to reach the battery potential the current is reduced to considerably low values. So obviously bulb won't glow. Note:As capacitor gains more and more charge its potential increases finally reaching the value same as the battery.So potential across bulb tends to zero. ...

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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 ...

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Suresh's answer gives the correct general formalism. (1) For the specific case of a coaxial cable, the electric field between the two conductors is determined by the charge $-Q$ on the inner conductor, which terminates on $+Q$ worth of charge on the outer conductor. (There can't be any field inside the inner conductor, so all the field generated by its ...

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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, ...

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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 ...

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I'll give you a simple answer. When you consider the capacitor as two parallel plate capacitors connected in parallel, you will see that their potential differences must be same. But not their charge. The charges on the two capacitors will be different. Thus electric field outside of dielectric in lower part of capacitor is not equal to the electric field ...

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Suppose on the other hand the field in the two places were not equal. Consider a loop integral around the red loop in say anticlockwise direction as shown in the figure. Only the vertical edges contribute to the integral.If $E_1 \neq E_2$,it is obvious that the loop integral is non-zero.This violates the conservative nature of the $\vec{E}$ field in ...

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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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If you just plug in your suggested solution, you get $$\frac d{dt} A\cos(\omega t + \phi)+\frac 1{\tau}A\cos(\omega t + \phi)=\frac{V_{in}}\tau\sin(\omega t)\\ -A\omega \sin(\omega t + \phi)+\frac 1{\tau}A\cos(\omega t + \phi)=\frac{V_{in}}\tau\sin(\omega t)$$ Now you should be able to use the function sum formulas to solve for $\phi$ and $\frac A{V_{in}} ... 1 Initially when you attach the capacitor to the battery, said battery will act to create an electric field within the wire. On the side of the negative terminal this field will point perpendicular to the cross section of the wire toward the terminal of the battery (electric field points toward negative charge). On the side of the positive terminal the field ... 0 Rather than saying that capacitance is the capacity to store charge you should consider it as the capacity to store energy. This lends itself well by writing the energy stored in a capacitor as:$U = \frac{1}{2} C V ^2$. If you were to keep the charge constant on a capacitor, you can't have freely moving charges, so the capacitor wouldn't have a current ... 0 Well that's the same doubt people face when they say that heat dissipated is proportional to R (I*I*R) as well as inversely proportional to R (V*V/R) My answer is that when you vary potential the charge WILL change unless you change the capacitance. If you have to keep the charge constant you have to vary the capacitance(i.e. separating parallel plates in ... 0 2) would be antiproportional instead 0 Capacitance is a measure of how much charge is required to make a change in voltage: $$C = \frac{Q}{V}$$ As the plates of a capacitor are brought closer together, capacitance increases. This is because the opposite charges on each plate of the capacitor can get closer to each other, and thus cancel each other more completely, and thus the voltage per ... 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 ... 0 Capacitance is defined as$C\equiv Q/V$. An interpretation of this is the following: Capacitance is the amount of additional charge stored on each plate for every unit of voltage increase across the capacitor. Capacitance gives you a sense of how much charge you get when you apply some set voltage across the terminals. A high capacitance means you get ... 0 Capacators are basically just parallel plates, so if you have them in parallel, then they act like two large plates (positive and negative terminals) and so can store more charge between them. If they are in series and the voltage drop between the first and last capacitor doesn't change, then the amount of charge you can effecively store in each capacitor ... 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 ...

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A single conductor also possess capacity to store charge. It may be treated as parallel plate capacitor, whose one plate is at infinity. If this doesn't help, comment on the part where you have problem.

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Okay, here's a simple explanation: You can consider a non-parallel plate capacitor as multiple parallel plate conductors, connected in parallel, with the distance between them increasing slightly for each capacitor. You get the net capacitance as the average of the minimum capacitance (based on the minimum distance b/w the plates) and maximum capcitance ...

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Here's what I do (or rather, what I just did) to convince myself of your result. We're trying to "reduce" the two capacitors in series to an equivalent capacitor. Equivalent in every way, including how much charge would flow when discharging the capacitor. Imagine discharging the two capacitors that are series. How much total charge would flow? Since, as ...

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Q here is the total charge that flows to the circuit from battery, In this case this charge is equal to the charges on the capacitors. If battery sends a charge Q then a plate of one capacitor gets Q. Thus the other plate gets -Q.By conservation of charge the capacitor connected to this plate acquires +Q and so on. Thus we get charges to be same on both ...

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