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First of all, the photon picture isn't the best way to visualize and model the operation of a radio-frequency antenna. RF antenna engineering is best performed and understood (IMHO) by means of a model in which high frequency AC power flow through a wire creates electromagnetic waves that propagate away into space. For this to occur, the antenna impedance ...


5

Remember that the word "capacitance" is just that, a word. The standard definition of capacitance given in introductory textbooks only applies when the two objects have charges $Q$ and $-Q$, in which case we define $$C = \frac{Q}{V}$$ where $V$ is the potential difference between the objects. There is no inherently "correct" way to ...


4

I thought however that since voltage is defined as energy/coulomb, the voltage across a capacitor is independent of the total number of electrons. The voltage $V$ across the capacitor of capacitance $C$ is not independent of the amount of charge $Q$ on the capacitor. The relationship between them is $$Q=CV$$ So for a given capacitance, the greater the ...


2

This is an interesting question but is asked in a rather vague way as the context given is limited. The standard definition of capacitance $C=\frac QV$ is useful because in many examples the capacitance is a constant and so $Q \propto V$. However there is the possibility of defining an incremental capacitance $C=\frac {\Delta Q}{\Delta V}$ when the charge ...


2

All elements of a circuit will radiate when there is alternating current involved. Typically though, at the frequencies involved, not very much. It is something that circuit designers must be careful of though. Capacitors themselves typically are encased in some sort of material which will damp the stray radiation, especially if it is a conductive material ...


2

How does the grounding of the negatively charged wire affect the charge distribution of the circuit/earth system and why doesn't the earths huge capacitance imply that a huge amount of charge must be required to charge it up to the potential of the negative wire? You have this backwards. The tiny amount of charge on the wire goes to the earth to bring the ...


2

To answer your question: The way you have written it, $V = E\, d$ is the potential due to both disks of the capacitor. When considering the force on just one of the disks, however, you must take half of that potential, since each disk as a whole does not "feel" its own electric field. In short, with your method, you must use $V = E\,d/2$. This is ...


2

This is a frequently asked question by college physics students. Since the situation in the question deviates from the usual definition and use of capacitors, we will need to generalize our definition of capacitance somewhat. Here, a useful generalization is the differential capacitance. Differential capacitance is the derivative, $C = dQ/dV$, and is ...


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Charge as a noun, the property of matter that is responsible for electrical phenomena, existing in a positive or negative form. the quantity of matter responsible for electrical phenomena carried by a body. energy stored chemically in a battery for conversion into electricity. an act or period of storing electrical energy in a battery. Charge as a verb, ...


2

It is true that voltage is energy per charge, but by "charge" it is not that ones accumulated on the capacitor plates. It means the work done on a test charge placed close to one of the charged plates, and being accelerated by the field to the other side. Of course electrostatic force is bigger for bigger charges, and also the work. So, to get a ...


1

The formula $E=\frac{k Q}{r^2}$ is the electric field of a point charge in radial direction at a distance $r$ from the charge. This means, especially, that the field is not homogeneous, i.e. not the same at every point in space. Compare this to your "the magnitude of the electric field is" and you'll find that we are either missing information or ...


1

The formula $$E=\frac{kQ}{r^2}$$ should be used for point charges or objects with spherical symmetry and uniform charge distribution. Using it for arbitrarily shaped uniform (or non-uniform) charge distributions will not allways give the right answer since the question as to what value you would use for $r$ will not be clearly defined. And I think that is ...


1

I guess you know how to calculate the output with a sinusoidal input wave. There are several techcnics for that, but the easiest one is probably using phasors. And thanks to Fourier series you can write your periodic rectangular wave input as a sum of sinusoidal waves. Let's take for example a square wave $V_{in}(t)$ with periodicity T: $ V_{in}(t) = \left\{ ...


1

In fact, you can charge electrons into a device, but that is not what people call a battery. Instead, you would be constructing an electrostatic capacitance in this way. These can produce high voltages, but have very limited energy density. In the popular battery no charges are added during the charging process. Rather, you pump potential energy into the ...


1

In simple metaphor, it's just like getting water into a ballon from a faucet. regardless of water pressure of the faucet, the pressure of the rubber shell is zero, However it goes higher as it accumulates. Just like so, Voltage across the capacitor is zero, without electrons accumulated. electron and electron hole is what creates the voltage between the ...


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I think your question might be valid in some non-technical sense. The truth is, the signal does lose energy. As someone previously stated, it is the power of the signal and the impedance of the atmosphere that matter. As the broadcaster of the signal, you have to push enough power through the air to send the signal. If you want to think of these bodies as ...


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I'm assuming the title of your post is your question. Please note that we can not provide solutions to homework and exercise type questions, only offer guidance. Here is some guidance: In a dc circuit under steady state conditions ideal capacitors look like open circuits unless shorted (as by switch S) and immediately after a switching event the voltages on ...


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