Energy in Capacitors

[rjosh33]

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Capacitors A and B are identical. Capacitor A is charged so it stores 4 J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now:

A. 16 J
B. 8 J
C. 4 J
D. 2 J

Answer: D

I feel like I'm missing something that should be obvious. Can anybody explain to me the answer?

 

[hiaips]

Capacitors A and B are identical. Capacitor A is charged so it stores 4 J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now:

A. 16 J
B. 8 J
C. 4 J
D. 2 J

I feel like I'm missing something that should be obvious. Can anybody explain to me the answer?

Well, the question is vague and could be interpreted in two perfectly valid (yet different) ways. Do they mean the energy stored in both capacitors together? If so, the answer is 4 J. (The energy doesn't change just by hooking up an extra capacitor.) If they mean the energy stored in each capacitor separately, then the answer is 2 J. Based on the answer provided, I am assuming that they are asking the latter question.

 

[rjosh33]

Well, the question is vague and could be interpreted in two perfectly valid (yet different) ways. Do they mean the energy stored in both capacitors together? If so, the answer is 4 J. (The energy doesn't change just by hooking up an extra capacitor.) If they mean the energy stored in each capacitor separately, then the answer is 2 J. Based on the answer provided, I am assuming that they are asking the latter question.

Yeah, I'm not getting it either, though I approached it a little differently. I tried using U = 1/2CV^2 = 4. With the new equivalent capacitance being twice the capacitance of A (capacitors in parallel are added), I then get
U = 2(1/2CV^2) = 2(4) = 8 J. Yet that's not right for some reason. Not sure what I'm doing wrong.

 

[N1st]

Yeah, I'm not getting it either, though I approached it a little differently. I tried using U = 1/2CV^2 = 4. With the new equivalent capacitance being twice the capacitance of A (capacitors in parallel are added), I then get
U = 2(1/2CV^2) = 2(4) = 8 J. Yet that's not right for some reason. Not sure what I'm doing wrong.

The second capacitor is initially not charged. Capacitance is defined as Q/V, so if you haven't charged the capacitor, the capacitance is 0. After you connect another capacitor in parallel, the charge just distributes itself equally, half to one capacitor and half to the other.

 

[rjosh33]

The second capacitor is initially not charged. Capacitance is defined as Q/V, so if you haven't charged the capacitor, the capacitance is 0. After you connect another capacitor in parallel, the charge just distributes itself equally, half to one capacitor and half to the other.

Ahhh, thank you. That makes sense. This problem had been bothering me for a couple days; glad to see it resolved. 🙂

 

[riseNshine]

so is the answer 4 or 8?

 

[milski]

The answer is 2 J in each capacitor, 4 J total.

 

[Postictal Raiden]

The answer is 2 J in each capacitor, 4 J total.

Is this because the energy is conserved?

 

[milski]

Yes, provided that the capacitors are not connected to the circuit that initially charged the first one but just to each other. It's hard to say whether that's really the case with how the question is phrased.