How do I calculate the total stored energy when capacitors are connected?

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To calculate the total stored energy when capacitors are connected, first determine the initial energy stored in the charged 2.21 µF capacitor using the formula E = 0.5 QV. After connecting it to the uncharged 3.60 µF capacitor, the total energy must account for the redistribution of charge between the two capacitors. The voltage across both capacitors will change upon connection, affecting the total stored energy. The final answer for the total stored energy after connection is 0.000168 J, which reflects the conservation of energy principle, assuming no losses in the system. Understanding the change in voltage and charge distribution is crucial for solving similar problems.
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Homework Statement


hey


A 2.21 uF capacitor is charged by a 20.0 V battery. It is disconnected from the battery and then connected to an uncharged 3.60 uF capacitor (Fig. 17-29).
http://www.webassign.net/giancoli5/17_28.gif

Determine the total stored energy at the following points in time.
(b) after they are connected


Homework Equations


C= Q/V
E=.5 QV

The Attempt at a Solution


I've tried to use the equations but I keep getting them wrong...
The answers I've gotten are
.00116
7.2e-4
.001279
.001162
4.42e-4
2.905e-4
884e-6

Thanks
 
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what points in time? also at which capacitor? or or both?
 
oh
after the two capacitors are connected
the first time was before they are connected.. which was 4.42e-4
I'm not sure .. but I assume it's both.. because that's all it says .. there's only one blank.. so I guess one answer.. >_<
 
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ok the question is not that clear to me. but a few things for you to think about: conservation of charge/energy, compare the voltage across C1 and C2think about what would happen upon connecting C2 to the system note C1 not eq to C2
 
I keep getting 4.42x10-6 ... but that's not the answer... TT________TT
 
if it is not clear what the question is really asking, it is very hard for me to help. what 's the correct answer?
 
I don't know.. It only tells me if I'm wrong..

the whole problem looks like this, exactly

"A 2.21 µF capacitor is charged by a 20.0 V battery. It is disconnected from the battery and then connected to an uncharged 3.60 µF capacitor (Fig. 17-29).

http://www.webassign.net/giancoli5/17_28.gif
Figure 17-29

Determine the total stored energy at the following points in time.
(a) before the two capacitors are connected
_____J
(b) after they are connected
_____J
(c) What is the change in energy?
_____J"

I think the V will change...
 
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my guess is that since it is asking for "total" stored energy ... assuming no loss in the wires, there should be no change at all. NB: 3.6uF capacitor is not charged, ie. does n't store any energy initially
 
thanks for helping... it for some reason.. it's not the answer..

can anyone else please try and help! TT_TT
 
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  • #10
hey guys..
the answer is 0.000168
but i don't know how.. >_<
can someone explain? thanks
 
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