1. The problem statement, all variables and given/known data A capacitor has been charged by connecting it to a 24 V battery. How much work was done in charging this capacitor, when at the end of charging, its charge is 3 µC? a. 8 µJ b. 12 µJ c. 36 µJ *d. 74 µJ 2. Relevant equations W=E=Q^{2}/2C=QV/2 3. The attempt at a solution W=E=QV/2=24V*3μC/2=36μJ The answer key says the answer is really 74μJ, but I dont know what I did wrong.
the answer you got is the energy stored on the capacitor. Can you think of any reason why the 'work done' does not all appear on the capacitor? What does the work?
what is the difference? Isnt the work done to charge it supposed stay on the capacitor as energy? Where does the rest go?
The battery voltage is 24V which means that each coulomb that comes from the battery gains 24Joules of energy (Volt means Joules per coulomb) So how much energy was given to the charge? The problem is accounting for the difference....IT IS NOT AN EASY QUESTION TO ANSWER How much work have you done on capacitors, what level are you at?
Sorry to break in, but I believe the attempt of physgrl is correct and the answer key is wrong. The first coulomb has to bridge a voltage difference of 0 V. When the capacitor has been charged with a charge q, the voltage difference that has to be bridged is V=q/C. In effect the work done in charging the capacitor is indeed W=QV/2 as you can also see on the wiki page: http://en.wikipedia.org/wiki/Capacitance
If there are 3uC then the voltage source gives 24V per C so it is really 24*3uC=72uJ ?? I took an ECE class so I have dealt with capacitors in term of their charge vs time and in circuits, but in this class weve done CV=Q C=C=εoA/2d and U=QV/2 (and the other 2 of those)
No, your original calculation was correct: W = 24V * 3uC / 2 = 36uJ As I said before, the 2nd formula should be C=εA/d, but yes, those are correct formulas. I do not know what technician intended, so I guess he should explain himself when (or if) he gets around to it.
The best answer to this question is (c)....no doubt about it. That is the energy stored on the capacitor so that is 'the work done charging the capacitor'.... that is all that the Wiki reference tells you. The fact that (d) was put forward as the correct answer made me doubt what they meant by 'the work done'. I still think (d) is wrong.... if it was 72μJ I would know what they were after. These are the facts as I see them 1) 3μC of charge come from a 24V battery, the WORK DONE BY THE BATTERY = 3 x 24 = 72μJ (Wikipedia makes no reference to work done by the battery) 2) the energy stored on the capacitor = 0.5V x Q = 36μJ (the best answer) The difference is energy lost during the transfer of charge.... any resistance in the circuit + electromagnetic radiation. Hope this helps
The first coulomb does not add much energy to the capacitor, because PD is low ≈0 the current is large and 'lost energy' (I^2R and radiation) will be 'high' When C is almost charged each coulomb increases the energy stored by a small amount, the current will decrease and the rate of energy 'lost' will decrease. ALL of the energy comes from the battery which has to 'lift' each coulomb through 24V (the emf) This is another example of conservation of energy in physics
I've been puzzling on it and I learned something new. :) The work done by the battery would indeed be: $$W_{battery} = \int V \cdot Idt = \int_0^Q V dq = QV$$ So it turns out than an energy of ##\frac 1 2 QV## is lost during the charging (as heat and/or radiation), regardless of ideal circumstances! I'm concluding that the energy would be burnt in the resistance of the wire (however small) and in the internal resistance of the battery (both cannot be zero at the same time). The question is indeed ambiguous. The work done by the battery in charging the capacitor is 24 x 3=72 uJ, and the work done on the capacitor in charging it is 24 x 3 / 2=36 uJ. Luckily 72 uJ is not one of the answers, so it must still be (c) 36 uJ.
ILS..I value your post which confirms what I contributed. Where would the wasted energy go IF there was no resistance in the wires and no internal resistance....there must be a physics explanation. I say it goes as electro-magnetic radiation....... I am certain that I have detected, in front of my students, this radiation using a pickup coil of wire and an oscilloscope. This is another bad, ambiguous question, but at least it has highlighted some important aspects of physics that needed clearing up I hope that physgrl who made the original post, can get some value from everything that has been posted.
I believe that with no resistance in the wires and no internal resistance, the circuit is a mathematical impossibility. The voltage across cannot be both 0V and 24V at the same time. The only way to make it work physically, is to assume a resistor or inductor somewhere however small. With a very small resistance, it would generate a lot of heat, which would show up as heat radiation (for a very short time). EDIT: Or with inductance, it would induce a magnetic field, which I presume is what you measured (the circuit does form a loop).
What about inductance , it is possible to specify zero resistance but still specify an inductance. I know this is way off the original post and it may be wrong/ misleading to raise it here. If it is accepted that there is bound to be inductance in the circuit then this will determine the rate of rise of current (rate of transfer of charge) and it is possible for an induced emf of 24 V to oppose the applied emf. I would genuinely like to hear your opinion but I do not want physgrls original post to become way off the original question. Cheers Tech
I had already edited my previous post to include induction. ;) Especially since the circuit does form a loop, which would induce a magnetic field.
Do not forget what happens inside the battery during charging the capacitor. There are chemical reactions. The energy of the battery decreases more than the increase of the energy of the capacitor. Part of "chemical work" transforms into heat - inside the battery. If you short-circuit a battery it will warm up. ehild
I found this on the hyperphysics site: From the definition of voltage as the energy per unit charge, one might expect that the energy stored on this ideal capacitor would be just QV. That is, all the work done on the charge in moving it from one plate to the other would appear as energy stored. But in fact, the expression above shows that just half of that work appears as energy stored in the capacitor. For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor. So I guess that half the energy will be lost in the wire, since it has finite resistance. (Unless maybe the wire is superconducting, but then there will still be the internal resistance of the battery).