Theoretical question about capacitance

[thisisfudd]

OK, let's say you have identical capacitors, and you charge one and do not charge the other. Then you connect the two. Will the energy in the two-capacitor system be the same as the energy in the first capacitor before it was connected?

 

[vincentchan]

no the energy will less...

so your next question is: where does the energy go?
resistance and radiation

 

[thisisfudd]

How do you quantify that?

 

[vincentchan]

energy store in capacitor is $$E=\frac{1}{2}\frac{Q^2}{C}$$
do the math

 

[Davorak]

There will only be energy loss if the capacitors are not considered ideal capacitors.
If we are talking about ideal capacitors there will be not energy loss.
Radiation loss is usually ignored as well in simple circuits.

However in the real case there is both loss to radiation and resistance just like vincentchan said.

Real capacitors are modeled by having an ideal capacitor in parallel with a resistor. There would also be some inductance if you put the two capacitors together.

This could be modeled by:
Node element node
1 C1 2
1 R1 2
2 C2 3
2 R2 3
3 L1 1

If you know some circuit analysis or have PSPICE handy you can find the losses through resistance fairly easily if neglecting radiation loss.

There may be some trick to getting a analytical answer to the radiation loss problem, but I do not know what it is. You could add the radiation losses equations to the circuit equation, but I have never solved a problem like that. I would go for a numerical solution here.

 

[thisisfudd]

OK, here's a question though. If E = Q^2 / C for the first part, and then

E = Q^2 / (2C) when there are two capacitors now, isn't it true that E is decreasing by a factor of two?

 

[Davorak]

Energy equation for capacitors:
$$E= \frac{1}{2}CV^2$$
$$C = \frac{Q}{V}$$
so
$$E= \frac{1}{2}CV^2 =\frac{1}{2} QV = \frac{1}{2} \frac{Q^2}{C}$$
In the Ideal capacitor case the charge divides among both capacitor equally since they are both the same capacitance. The energy in the charge capacitor before connecting it to the other capacitor is:
$$E= \frac{1}{2} \frac{Q^2}{C}$$

How does the charge divide up between the two capacitors?

Are the capacitors in series or parallel?

How do both of these considerations effect the equation for energy?

 

[thisisfudd]

We haven't really learned the series and parallel thing yet in class so I am assuming we don't have to apply that, I hope. I know that E = 1/2 Q^2/C. But when it divides up between the two capacitors (evenly) there are still 2 capacitors, so there is 2C. So then:

E = 1/2 Q^2/2C or 1/4 Q^2/C, or 1/2 of what it was before.

 

[vincentchan]

yes, the energy is one half of that b4...

There will only be energy loss if the capacitors are not considered ideal capacitors.
If we are talking about ideal capacitors there will be not energy loss.
Radiation loss is usually ignored as well in simple circuits.

In this problem, if you are asking the final equalibrium steady state, you cannot assume no energy loss in the system... this problem is kind of similar to a spring mass problem in homonic osccilation.. if no energy loss (damping), the system will osccilate back and forth and never stop (that's means the charge will go from capacitor A to B, then go back to A, and B, and forever)

 

[Davorak]

In this problem, if you are asking the final equalibrium steady state, you cannot assume no energy loss in the system... this problem is kind of similar to a spring mass problem in homonic osccilation.. if no energy loss (damping), the system will osccilate back and forth and never stop (that's means the charge will go from capacitor A to B, then go back to A, and B, and forever)

What vincentchan is correct. For when the two capacitors are connected together:
$$E_{Before}=E_{after}= \frac{1}{2}\frac{Q_1^2}{C_1} + \frac{1}{2}\frac{Q_2^2}{C_2} +KE_{electrons}$$

Does this make sence?