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There 3 rather simple formulas which allow to compute the parameters that are asked for. $L'=\mu/(2\pi) \cdot \ln(b/a)$ for the inductivity per meter L' $C'=2\pi\epsilon/(\ln(b/a))$ for the capacity per meter C' $Z= Z_0/2\pi \sqrt{\frac{\mu'}{\epsilon'}}\ln(b/a)$. b is the outer diameter and a is the inner diameter of the coaxial line $Z_0 = ...


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There is only one connection to the wrapper, at the power supply end. One can read the problem to imply that the resistor at the other end is connected to the wrapper, but you shouldn't. If there were a connection there, the resistor would be shorted out. As the wrapper has only one connection, at long $t$ it will have come to the proper equilibrium ...


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A capacitor is assumed to be self-contained and isolated, with no net electric charge and no influence from any external electric field. The conductors thus hold equal and opposite charges on their facing surfaces. As electric field is established which originate from the positive plate and end on the negative plate. Also, the field is uniform so is the ...


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I suppose that in this type of circuit the current and the voltage are in phase, This question is asking for a transient solution, not an ac steady-state solution. So we can't really talk about the phase of the voltage or the current. meaning that immediately after the switch is shut the current should be I=0? This conclusion is incorrect. ...


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Another way to look at this is to realize that whenever you move charge, you produce a magnetic field. Even a straight wire will have a certain amount of self-inductance. Initially, this will provide a limit as to how fast the current can increase. As the voltages equalize, the interaction with the magnetic field will actually cause the current to continue ...


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Even an ideal capacitor cannot be losslessly charged to a potential E from a potential E without using a voltage "converter" which accepts energy at Vin and delivers it to the capacitor at Vcap_current. If you connect an ideal voltage source via a lossless switch to an ideal capacitor which is charged to a lower voltage, infinite current will flow when the ...


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Consider the three-terminal device that is your stacked capacitor: A ----============================================= (dielectric medium ɛ) =============================================---- B (dielectric medium ɛ) C ----============================================= All three plates have the same area A. It's ...


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Yes it is right to say that electric field by a finite plate is not constant and zero at infinity. But in case of capacitors,the separation between the plates is so small as compared to dimensions of plate that with respect to the separation between the plates the plate itself can be considered as infinite. It is just a relative assumption to simplify ...


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Potential depends on the reference point,in most cases we take the reference point to be at infinity,i.e potential is zero at infinity.(its our choice). But if you try to take the reference point at infinity for a infinite charge plate the equations blow up.Thats why we cannot take the reference point to be at infinity for such plates. But we can then take ...


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The charge on the two plates will get distributed in such a manner: (+1C)|||(+3C) (-3C)|||(+1C). There is a particular behaviour that I noticed(while doing many questions of the same type) in which the charges on the outer surface of the plates is equal to the sum of all the charges present on the combination of plates... This is what I have used ...


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It's not entirely clear from your question, but it appears you have 6 coulombs charge on a capacitor C, which is then distributed to be on a capacitor 2C. The rest should be obvious application of the voltage on a capacitor equation.


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Consider the following cases in relation to your question: Inner sphere is grounded. a) grounding the outer surface of the inner sphere If you ground the outer surface of the inner sphere, the inner sphere becomes irrelevant and you get single spherical capacitor (the other one at infinity) of radius b. The capacitance is now $4\pi\epsilon_{o}b$. b) ...


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Only some circuits can have such a representation. Such circuits can be represented as a resistor in series with an inductor in series with a capacitor. This is hardly the only circuit possible. For instance, an electrolytic capacitor is sometimes (depending on the use being considered) represented as a cascade of resistors and capacitors, with each ...


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Good questioning. I don't know for sure what eventually lead us to adopt the paradigm of RLC circuits, but do know that the physics of electrical circuits deals with the way energy flows. In a circuit energy can be stored in the form of magnetic or electrical fields, and so the inductor and capacitor manage that capability. Energy can also be lost, and thus ...


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Tom's answer is completely correct, but I wanted to add some detail to it. First, why is it that your logic fails? The answer is that the familiar formula for the capacitance of two parallel plates relies on the approximation that the electric field between the two plates is completely uniform. In such a case, we have $E = V/d$ and $E = Q/\epsilon_0 A$; ...


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Consider the case of a conducting sphere with no counter electrode. The capacitance of the conducting sphere can be found with $$C=Q/V$$ and we know that at a distance $r$ from the centre of the sphere the potential $V$ is given by $$V={1 \over 4\pi\epsilon_0}{Q\over r}$$ provided that $r$ is greater than the radius of the sphere. Now if we consider ...



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