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I was looking at the calculation of the energy stored in the capacitor, and I don't see why the calculations make sense. It goes as follows: the potential difference and charge on a capacitor satisfy the equation $$V=\dfrac{Q}{C}$$ where $C$ is the capacitance. Suppose a charge $dQ$ is taken from one plate to the other. The work done to transfer the charge is $$dU = VdQ$$ which implies that $$dU = \dfrac{Q}{C}dQ$$ and then the integral is taken from $0$ to the final charge $Q$(source: Feynman Lectures).

Now, I know that a capacitor is not charged by taking some charge from one plate to the other. Rather, it is charged because of the electric field of a battery of the circuit in which the capacitor is connected: the electrons move from the +ve plate of the capacitor to the positive terminal of the battery, thus making it positively charged, and an equal number of electrons move to the -ve plate of the battery, thus making it negatively charged. So how is the work done calculated this way correct?

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What the battery does is to intake some negative charge from the positive terminal, transport it through the battery against the electric field (here the work of chemical forces against electric forces happens) and outputs equal charge of positive sign on the negative terminal.

The larger the capacitance of objects connected to the terminals (capacitor plates), the larger the charge transported. As the charge is transported, work is being done by the battery to create this macroscopic separation of accumulated charge. Part of this work gets stored in the capacitor and can be expressed either as $\frac{1}{2}CV^2$ or as $\frac{1}{2}Q^2/C$.

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  • $\begingroup$ Pedantic comment: the energy stored in the capacitor is, as you write, $\frac{1}{2}CV^2$ but the work done by the battery in charging the capacitor is (ideally) larger by a factor of 2. $\endgroup$ – Alfred Centauri Apr 24 '18 at 12:02
  • $\begingroup$ @AlfredCentauri, you're right, the charging isn't ideal reversible process. I've edited my answer to correct that. $\endgroup$ – Ján Lalinský Apr 24 '18 at 13:48
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Now, I know that a capacitor is not charged by taking some charge from one plate to the other. Rather, it is charged because of the electric field of a battery of the circuit in which the capacitor is connected:

A battery isn't the only way to charge a capacitor is it? Consider the charging of a capacitor by an inductor rather than a battery. For example, look at the following circuit:

enter image description here

Initially, the switch is in the left-hand position as shown, the inductor current is constant and the voltage across the inductor is zero.

When the switch is thrown to the right-hand position, charge flows from one plate of the capacitor to the other through the inductor and the so the inductor begins charging the capacitor via the energy stored in the magnetic field of the inductor.

Importantly, at the instant after the switch is thrown, there is no voltage across the inductor (or capacitor); the inductor is at this moment acting as a current source that is 'pumping' charge from one plate of the capacitor to the other.

Indeed, here's another way to charge a capacitor sans battery:

enter image description here

The current source charges the capacitor, until the voltage across the capacitor equals $V_C = I_1 \cdot R_2$, by moving charge from one plate of the capacitor to the other and, in the process, doing work against the building voltage across the capacitor.

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  • $\begingroup$ how does this answer the question of calculating the energy stored in the cap? $\endgroup$ – ZeroTheHero Apr 23 '18 at 22:15
  • $\begingroup$ Hi @ZeroTheHero. In my answer, I've attempted to address what I believe is a key misconception in the OP's question - one that if corrected might make the question of calculating the energy moot. I've highlighted the alleged misconception in the quote. $\endgroup$ – Hal Hollis Apr 23 '18 at 23:41
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A charge disappears from the positive plate. A charge appears on the negative plate. Is your concern that the charge that disappears from the positive plate may not be the "same" as the one that appears on the negative plate? Subatomic particles are indistinguishable. There's no difference between a charge moving from the positive plate to the negative plate, versus one charge disappearing, and a "different" one appearing. Certainly, the amount of work needed is the same.

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