Energy loss in capacitators

[ronthepon]

I have recently studied capacitators and came across this little result.

 

Suppose we have two capacitators, each charged to some extent.

 

Capacitator1:

Capacitance- [math]C_1[/math]

Charge- [math]Q_1[/math]

 

Capacitator2:

Capacitance- [math]C_2[/math]

Charge- [math]Q_2[/math]

 

Now energies of both will be = [math] \frac{{Q_1}^2}{2{C_1}} [/math] and [math] \frac{{Q_2}^2}{2{C_2}}[/math]

 

The total energy of the system will be the sum of both, [math] \frac{{Q_1}^2}{2{C_1}} + \frac{{Q_2}^2}{2{C_2}}[/math]

 

Now, let them be connected to each other.

 

The new capacitance will be equal to [math] C_1 + C_2 [/math]

 

And the total charge in this combo-capacitator will be [math] Q_1 + Q_2 [/math].

 

Energy stored will be [math] \frac{{Q_1 + Q_2}^2}{2(C_1 + C_2)} [/math]

 

[math] = \frac{{{Q_1}^2}+{{Q_2}^2}+{2{Q_1}{Q_2}}}{2{C_1}+ 2{C_2}}[/math]

 

There is an energy difference of [math] {\frac{{C_1}{C_2}}{{2{C_1}+ 2{C_2}}} ({\frac{Q_1}{C_1} - \frac{Q_2}{C_2}})^2[/math] (Take my word its positive and finite)

 

Where does this extra energy go?

[Qfwfq]

It is partly dissipated and partly radiated.

 

BTW, if you look carefully, the difference may be zero.

[ronthepon]

No, the difference is not zero, I tried substituion at many times.

 

I was given the radiation and dissipated answer by my seniors, and they absolutely hate being questioned.

 

From exactly where is it radiated? Does it become heat energy? And why?

[cwes99_03]

How do you have these capacitors connected. It has been 6 years since I worked with these equations, but I am quite certain that there are two different equations depending upon whether they are connected in series or parallel.

[Tim_Lou]

when you connect the capacitors, the electrons move from one plate to another, generating current. since the resistance of the wire you use is always non-zero, hence there will be heat dissipated or radiation generated the instance when they were connected.

[Erasmus00]

The combined capacitance is only C1+C2 if you connect them in parallel. In order to do this without driving currents, the two capacitors must be at the same voltage before you connect them in parallel. This implies Q1/C1 = Q2/C2 and there is, in fact, no energy difference.

-Will

[cwes99_03]

Ok, so I went and found a resource.

 

http://www.oz.net/~coilgun/theory/capacitors2.htm

and

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html

 

From these two sites we gain that hooking

1) two capacitors in series gives us a total capacitance of the inverse of the sum of the inverses (someone want to put that in to TeX.)

2) two capacitors in parallel gives us a total capacitance of c1+c2.

 

Energy stored in a capacitor is not (Q^2/C) but 1/2(Q^2/C).

 

Care to try those calculations again?

[ronthepon]

I had considered paralell capacitators, I had used the 1/2 element.

 

And resistance of the wire is considered negligible, because there is no mention of which wire they are connected by, they could be connected by super-duper-conductors (with theoretical zero resistance, thus instant transfer of electrons).

[Qfwfq]

If you apply a force F to an object of mass m, in the vacuum so that there's no friction against the force, what happens Ron?

[ronthepon]

It accelerates? And follows the F=ma formula.

[sebbysteiny]

My suggestion is that when you connect the wires, you create a short circuit until the voltages are equal. As you know, whenever you get a short circuit, you get massive loss of energy (hense the heat in a fuse melts the fuse wire).

 

So the problem is the same as why do you lose heat in a short circuit.

 

I think the answer lies in the energy being used to accelarate electrons, and the electrons remain accelarated until they collide and give up all their energy in the form of heat. They will collide with the metallic latice and disipate energy, it's just a question of how quickly. In a super conductor, they will not collide, but the energy will be transfered instead to the kenetic energy of the electron (well technically, the cooper pair, he he).

 

Hope this helps.

[Qfwfq]

It accelerates? And follows the F=ma formula.

Of course! :(

 

And if the mass m also has a charge e, what happens when it accelerates?

[ronthepon]

And if the mass m also has a charge e, what happens when it accelerates?

 

I believe that the charged particles radiate energy

 

The acceleration of the electrons causes the energy loss?

[Qfwfq]

Exactly, :) this is the most fundamental way to put it.

 

Macroscopically, you can also say that, while there is such a thing as a superconductor, you could hardly get out of there being inductance when the two capacitors are connected. Even if you put the parallel plates in pairwise edge-on contact, the system would have an LC cell in its complete model. If, as a limiting case, you strove to reduce the radiation toward zero, what would the behaviour be like? Try writing the energy of the system, in this limit and assuming a value of L as well as of C. To simplfy, it's quite equivalent to having only one of the two capacitors and shorting its two terminals.

[Qfwfq]

P. S. I was meaning to say that the energy difference of

 

[math]{\frac{{C_1}{C_2}}{{2{C_1}+ 2{C_2}}} ({\frac{Q_1}{C_1} - \frac{Q_2}{C_2}})^2[/math] will be zero if and only if

 

[math]{{\frac{Q_1}{C_1} = \frac{Q_2}{C_2}}[/math]

 

which is the equal voltage case that Will pointed out. Sure, it can't be negative.

 

If your seniors resent being questioned, find better ones... :hihi: you might learn a lot more. :)

[ronthepon]

I get it now. Thanks.