# Tag Info

2

These are all the same equation, because $C = \frac{Q}{V}$. You can convert. As long as you know any two of $Q$, $C$, and $V$, you can use the equation that references those two, or use the formula above to convert to another one.

6

A battery connected to a capacitor is an RC circuit in the limit $R \to 0$ (i.e., there is no resistor and the resistance of the wire is negligible). One might think that the energy loss is zero in this limit, but this is not the case. For an RC circuit with a battery and an initially (i.e., at $t=0$) uncharged capacitor, we have Q(t) = CV ...

1

An ideal capacitor never "dissipates" energy, it merely stores it. The amount of energy stored in a capacitor is given by the formula you mentioned: $U = \frac{1}{2}CV^2$. In the case of the LC circuit, the energy stored in the capacitor moves into the inductor in form of magnetic field energy and then goes back and forth from them. In the case of an ideal ...

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Well for me it works that since for both charging and discharging, Q is inversely proportional (exponentially) to $t$, hence it is always decreasing in its decrease or increase with time. Hope it works for you too :)

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"My question is how are we getting the energy loss without taking into consideration any terms that cause it at all? Or have we taken it and I missed it?" You know, I was also puzzled by the very same question when I was presented with the fact that there is an energy loss in going from a single charged capacitor configuration to a configuration where the ...

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If you find the energy of the two configurations and subtract them you will see they are unequal. Firstly this makes sense, because if you brought two capacitors together you wouldn't expect all the charge to go to one capacitor, so that configuration with all the charge on one must have higher energy. Later, when you introduce resistors you will will see ...

2

Just to expand a little bit on Emilio's answer, and to address the specific point you made in your final paragraph about "finite charges", I decided to calculate what fraction of electrons would be "missing" from the positive plate of a capacitor when you reach breakdown. For a vacuum dielectric, the Schwinger limit is about $1.3\cdot 10^{18}\rm{V/m}$ - ...

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The charges that accumulate on the plates of a capacitor are not provided by the material of the plates themselves but by the source that is charging them, so there is in principle no limit to the amount of charge that they can hold, if your source is strong enough. The maximum-charge limits on actual physical capacitors are dictated by the dielectric ...

0

This site from the University of Texas provides a good explanation, I think. Basically, when you have two capacitors connected in series, say $C_1$ and $C_2$, then the total charge in the middle wiring connecting the two components must remain constant, as it cannot escape anywhere. Any charge accumulation in in $C_1$'s outer plate creates a virtual ...

1

Think of two capacitors in series. If electrons e.g. move to the left plate of capacitor $C_2$, that plate gains some negative net charge, let's call it $-Q$. The negative charge on the left plate repels like-signed charge on the right plate (electrons are repelled from the right plate and pushed away to the right). This right plate of $C_2$ thus gets a ...

1

Let's say we hook up a capacitor to a battery (and maybe toss a resistor in there). The battery will pump a charge difference between the plates, which creates a potential difference between the plates. When the potential difference reaches the potential difference of the battery, current that takes and adds charge to the plates stops flowing, as current ...

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The potential difference or voltage is actually the difference in the level/numbers of electrons or positive charges on the two terminals of battery or across the terminals of passive elements. Law of conservation of energy states that the total energy of an isolated system remains constant, that is why the sum of the potential differences across the passive ...

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To understand the answer, you need to be aware of the concept of electric potential. Electric potential is a scalar quantity. In any circuit, there is a potential at any given point on the wire. The difference in potential between any two points in this circuit is sometimes called potential difference or voltage. You can understand the difference between ...

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You have essentially answered your own question. You are correct that the potential as we call it is a measure of the amount of work that would be done per unit of charge as it is moved through that potential. Electric fields are conservative in that the amount of work that must be put into moving a particle from one potential to another is exactly equal ...

1

When thinking about inductors on a conceptual level, the thing to remember is that they oppose change in current. In other words, if the current, $i$, is dropping, they provide voltage in the direction of that current; if $i$ is increasing, they provide voltage in the other direction (this can be very loosely thought of as resembling "inertia" in the ...

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Let's say we have a capacitor with capacitance $C$ with stored charge $Q_0$ at an initial potential difference of $V_0$. Now we are going to discharge it say over a simple resistor and determine how much energy was stored. During discharge in a small time interval $\Delta t$ the energy $\Delta E$ dissipated from the capacitor will be: $\Delta E = V I ... 1 I have doubts about only one conducting plate. H&R say that the field outside is E=σ/2ε0, but if you add vectorially the two fields due to the two surfaces, then the field is E= σ/ε0. Inside the plate the vector addition is zero. When there is two plates the field inside the metal of the two plates, due to the four surfaces, is not zero and there must ... 2 Wikipedia defines capacitance of parallel plates as $$C=\frac{Q}{V},$$ where$\pm Q$is the charge on the plates (one sign for each plate) and$V$is the voltage between them. In other words, yes, if you calculate a capacitance to be negative from, say,$Q<0\$, then you can just take the magnitude and call the capacitance positive.

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