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0

Yes, the field outside the conducting surface is 0. If the net charge inside the conducting surface is 0, then the fields outside the surface are zero. It's basically the Faraday Cage effect. And the net charge inside the conducting surface is 0, because that's the definition of a capacitor. You might ask, "Um, well, but what if the net charge is not ...


1

The question is more about battery chemistry than physics, but here are some things to keep in mind: Capacitors can typically retain MUCH less charge than a battery, since the latter stores energy in chemical form Supercapacitors are a class of capacitor that can be used for precisely the purpose you describe. From the Wiki page: Supercapacitors are ...


0

My suspicion is that the output of your first capacitor still has a 2.8 V peak-to-peak 60 Hz ripple on it. There is always some level of ripple out of the usual rectifier + capacitor DC power supply. Putting a diode between the first and second capacitors creates a "peak detector" that samples the highest points of this ripple at +1.4 V above your 160 V ...


0

Connecting two perfect capacitors like that would be like connecting two perfect but different voltage sources; you would get a hypothetical explosion. In real life, every capacitor has inductance and resistance. So, as the current built up between the two capacitors, you'd heat up the wire between the capacitors as well as the capacitors themselves. If the ...


0

Displacement Current actually does not exist, it is a theoretical misnomer. When we consider a Capacitor as a low Characteristic Impedance Transmission Line we can think of energy flow between the conductors (Parallel Plates) We see a TEM wave (ExH) moving at the speed of light for the medium. What Maxwell thought of as Displacement current is the ...


0

If you are considering a fixed charge on the capacitor, it won't matter at all. The charges have plenty of time to distribute themselves evenly over the plates, so there is no current anywhere, so no voltage drop. If you are considering conditions during the charge cycle, the charges in the low resistance plate will spread out more quickly. If the leads ...


1

When you connect a battery to a capacitor, a "real" circuit has at least four components in series: the voltage source (battery) the capacitor series resistance series inductance Any wire has inductance, since current flowing through it will induce a magnetic field. It is possible to have wires without resistance - we call them superconductors. So let's ...


4

I understand also that there would be a tiny minuscule resistive loss through the wire, but really it's not enough to say anything about. On the contrary, it's crucial. Assuming an ideal voltage source (can supply unlimited current) of voltage $V_S$, an ideal resistor of resistance $R$, and an ideal uncharged capacitor of capacitance $C$, are ...


0

Here goes my answer :-) Being $(V_1,Q_1) \;\&\; (V_2,Q_2)$ the voltage and charge of each series cap $(C_1\;\&\; C_2)$, we know (KVL) that the voltage of the ensemble is: $$V_e=V_1+V_2. \tag{Eq.1}$$ and as you said (KCL), $$Q_1=Q_2 \tag{Eq. 2}$$ Where ${1\over2}\cdot Q_1\cdot V_1 \;\&\; {1\over2}\cdot Q_2\cdot V_2$ are the energy stored in $C_1 ...


0

I think your approach isn't wrong; however in your calculations you're making the assumption that the potential difference between plates, $V$, is constant: What remains constant is the charge on each plate. So the equation becomes: $$W=\int_0^d {{q^2} \over {2\epsilon_0A}} \;\mathrm{dx}={{q^2d} \over {2\epsilon_0A}}$$ Since $C=\epsilon_0A/d$ we obtain ...


0

From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d. That's not the ordinary understanding and, from the electrical circuit perspective, the energy stored equals the work done by the external circuit separating electric charge in the ...


0

The calculation is not right, but the result is true. You need infinite work to pull apart two oppositely charged infinite plates, they have each infinite charge! The electric field due to one infinite plate is $\frac{\sigma}{2 \epsilon_0}$. The force on an area $A$ of the second plate will thus be $\frac{- \sigma^2 A}{2 \epsilon_0}$, which is infinite for ...


0

As John Rennie marked, the result should be $\tfrac{1}{2}CV^2$. Let me deduce this for you; Let's start with an uncharged condenser & by some means you remove one an electron from one plate & transfer it to the other plate. You have to do hardly any work to transfer the first electron but as you gradually continue the process, the field that is ...


1

NO. The Capacitors are in series as when we go from one capacitor to another we find no junction in between.


1

No. The capacitors are in series. This is because one side is at the same potential. But it looks like parallel. If you apply Kirchoff's Loop Law, you will see that they are in series. And in that way too, you will get that answer. Hope that helps.


7

There is no analogy of "voltage", "current", "charge" or "flux" to electromagnetism for the weak force, at least none that would be helpful. The reason for this is that all of these are classical concepts, while the notion of the weak force is completely quantum. Taking the classical limit just makes it vanish because the classical force law of forces with ...


-8

Are there analogs to resistance, inductance, capacitance, and memristance connecting the weak force to electromagnetism? Yes, but not the way you're thinking. Take a look at weak interaction on Wikipedia and note this: "The first type is called the 'charged-current interaction' because it is mediated by particles that carry an electric charge (the W+ ...


2

You refer to plates, so I assume we are talking about a parallel plate capacitor. In this case (to a good approximation) the field is the same everywhere inside the capacitor so if the field is bigger than $E_b$ somewhere inside the capacitor then it is bigger than $E_b$ everywhere. But if the capacitor is not a parallel plate capacitor all bets are off. ...


0

When you connect the negative terminal to ground, why would the negative charge flow anywhere? It has to exist somewhere, and it might as well be close to the positive charge on the other capacitor plate. You could also consider a very similar scenario in which you connect the negative side to ground before disconnecting the charging source. In this case, ...


0

We have the surface charge density as charge/area. Now consider


0

This is how ideal and impossible circuits elements behave, but it's a starting point for a simple analysis: At discontinuous changes in circuits, 1) inductors have the same current immediately before and after the discontinuity, but can have discontinuous voltage changes. The current will then change exponentially/sinusoidally/both. 2) capacitors have the ...



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