conservation of charge r energy?

Thread Starter

chandraprakash

Joined Dec 28, 2006
2
10v
|
|
== 3f
|
|
___
"""""
a


|---||(3f)----|0v
|
|----||(6f)----|0v

what ll b the voltages across the 3f and 6f if 3f was connected to a 10v source initially..ll they follow conservation of charges or energy?...
 

Thread Starter

chandraprakash

Joined Dec 28, 2006
2
if u consider charge conservation u lla voltage of 10/3 across both the capacitors...but energy s not conserved..
1/2*3*10^2!=1/2*3*(10/3)^2+1/2*6*(10/3)^2...
 

n9352527

Joined Oct 14, 2005
1,198
There is missing energy associated with transfer of charge between a charged capacitor to an empty one. If you have two capacitors with the same value, and one is charged to say 10V and the other is empty, then upon connecting both of them in parallel, the final voltage across both of them is 5V. This is calculated through the charge conservation, i.e. Q = CV.

However, the total energy stored in the capacitors is now a half of the initial energy (from 0.5CV^2). The explanation is a bit sketchy, but the best I could find is the so called Lacy-McCabe law, where the potential of doing work (or in this case stored energy) can only be realised by doing work and not by passing the stored energy directly without changing form.

This is an old problem, which has so many explanations and approaches that it could end up more confusing for the reader. There are also explanations that the energy is dissipated in the transfer (electromagnetic, heat, etc).

Have a look at this paper http://ej.iop.org/links/rEIkex2sU/qH-lBlWa2xG04EyDav5vpA/pe7202.pdf
 

mrmeval

Joined Jun 30, 2006
833
That link is not there, it's to an article that has to be paid for and it's a temporary link or one only you have access to.
 

n9352527

Joined Oct 14, 2005
1,198
You are right. I apologise.

You could search for Lacy-McCabe Law pertaining to transfer of charge between two capacitors and conservation of energy instead.
 
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