New answers tagged

-1

Building upon the previous answer of mike stone. So, if a specific material medium modeled as a signal transmission line, insulated with a dielectric and assuming all the signal is on the surface of the line (i.e. high frequency signal skin effect) has $ε$ permittivity (i.e. also known dielectric constant) and permeability $μ$ specific values and also $C_{0}$...


2

Yes. If the transmission line is surrounded by air then $$ \frac{1}{\sqrt{\mu_0\epsilon_0}}= \frac{1}{\sqrt{LC}}. $$ You can compoute the value of a $L$ and $C$ for a pair of coaxial cylinders, for example, and see that this relation holds. If the transmision line is embedded in a dielectric then the same relation holds, but with $\mu_0$ and $\epsilon_0$ on ...


0

The actual capacitance is higher due to the edge effect because the field lines extend beyond the plates. See the following for a calculation https://www.calctown.com/calculators/edge-parallel Hope this helps


1

Note that in some definitions of electrical-to-mechanical analogies, the capacitive terms in each are reciprocals of one another. Might this be the source of your confusion?


0

Most textbook assume that plates are infinitely long so for infinitely long plates electric field is p/2e where p is charge density on plate and e is permittivity as you can see p is constant electric field independent of distance too remains constant.


0

Energy and mass are equivalent. If the electric field stores energy, then it stores mass. Weight is proportional to mass, other things being equal, so the electric field will contribute to the weight of the capacitor. Likewise a wound up mechanical watch has a greater weight than the same one unwound, other things (e.g. temperature) being equal. It's really ...


0

This was tested experimentally by Kreuzer, Phys. Rev. 169 (1968) 1007. The "capacitors" in this experiment were actually atomic nuclei. The electric field of a heavy nucleus makes a fairly significant contribution to its own weight, which is easy to measure. Kreuzer used a Cavendish balance to test whether this was also true for the active ...


-3

IMO No. The electromagnetic energy from the source is used to coherently rearrange the existing charges in the circuit. It does not add more matter or energy in this open system. The electric potential energy of the source is transformed in kinetic energy of the moving charges and all of this energy is expelled out of the system as heat. After the capacitor ...


7

Yes, the electric fields in a capacitor add to its weight. But not so that you’d notice with anything so crude as a balance. Suppose you had a one-farad “supercapacitor” that you could charge up to one kilovolt. The energy stored in the electric field would be $$ U = \frac12 C V^2 = \frac12\times10^6 \rm\,J $$ This is an awful lot of energy for a capacitor, ...


8

I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth. General relativity is not required to answer this question. Consider (for simplicity) a parallel plate capacitor where the field is constrained within the parallel plates. In this case the field is uniform (let it be $E_0$) ...


1

You should have trouble with it. It isn't true. For example, suppose $$C_3 = 2C_1$$ $$C_2 = 10C_1$$ $$C_4 = 20 C_1$$ You will find that $$V_{ac} = V_{ad}$$ But $$Q_{right} = 10 Q_{left}$$


0

In the above circuit $C_1$ and $C_3$ are in series and also parallel to series capacitors: $C_3$ and $C_4$. There is a voltage potential between $a$ and $b$, but none between $c$ and $d$. Capacitors in series have the same charge, and the charge on the left and right sides is the same. For the series capacitors: $$left \ hand \ side \ \ \frac{1}{C_{s1}} = \...


1

It is true, only if the capacitors have capacitance values that are symmetric ( $C_1 = C_2$ and $C_3 = C_4$ ). Else is not true. You can do the computation using the formula of a series of capacitors $1/C_{tot} = 1/C_1 + 1/C_2 +...+ 1/C_n$ (here $n = 2$). You can compute the charge by the well known relationship $Q = C \times V$.


0

This is how I look at capacitors. When the battery is connected electrons are pushed from the battery and accumulate on the capacitor, Just to be clear, the battery doesn't supply electrons. The battery does work to move free electrons in the circuit from one plate to the other, leaving the plate where electrons are removed with a net positive charge and ...


0

The voltage across the plates is equal to the battery in 2 scenarios: Where the system is in equilibrium When you're modeling an ideal scenario rather than a real one. The key to the first point is that, were the voltage to not be equal, there would be a voltage driving a current which charges the capacitor. The only time where there is no current is when ...


2

There are no critical points because this function doesn't have local minima! The function is linear in $L$. To maximize, we want to make $L$ as large as we can. Assuming $b>a$, the function $\frac{1}{\ln(b/a)}$ is monotonic decreasing. To maximize, we want the ratio $b/a$ to approach $1$ from the positive direction. To summarize, if there aren't any ...


0

Your second formula should be $V=Ed$


2

Calculations, based on conservation of charge, show that, for a charged capacitor, A, connected to an identical, initially uncharged, capacitor, B, the new voltage between the plates of A or B is half the original voltage between the plates of A. I assume that you can do calculations of this type, but are seeking a more intuitive reason for the voltage ...


0

If they are initially discharged, put in series, then charged this is true because of Kirchoff's current law. However, If you charged the capacitors separately then connected them in series this may not be true. Also, from your problem wouldn't introducing a dielectric part of the way between plates be modeled as two parallel capacitors (one with and one ...


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