Where Does Half of the Work Done by a Battery Go in a Capacitor Circuit?

[ukamle]

We know that energy stored in capacitor=

$$\int \frac{q}{C} dq = \frac{q^2}{2C} = qV/2$$

But work done by battery = qV

Where does the other qV/2 go ?

Assume NO resistance in circuit

The potential of the battery does not change with time

 

[ceptimus]

If the circuit really had zero resistance then the initial current would be limited by the circuit's inductance. And it would never reach a steady state - the voltage on the capacitor would oscillate between 0 and 2V, with energy being transferred back and forth between the capacitive and inductive elements.

The situation with no resistance and no inductance is not possible - it would lead to an infinite current flow, and the capacitor would charge in zero time.

 

[rayjohn01]

Cap Resistance

Unless a capacitor in totally enclosed it is by definition an open spaced device
i.e. there is a gap between the electrodes . As a cap charges there is a current in this gap called the displacement current -- such currents are the cause of or due to the fields which exist in the gap , and they can escape.
Any field energy which escapes can be modeled as resistace so that by definition a capacitor has resistance --- it can be very small but that only means that if the inductance approaches zero the the currents are very large so the effect is the same -- energy loss to space.
Plus Septimus is correct in that a cap has to occupy some spatial dimension ( with conductors ) any such arrangement will also have inductance so the net result is that any cap has both ind , and res except in the totally imaginary condition which then gives rise to the paradox.
Ray.

 

[reilly]

This is generally covered in freshman physics (see, for example, Halliday and Resnick, or the more advanced Smythe Static and Dynamic Electricity). The easiest way for you to solve your problem is to start with a battery-capacitor system, with the capacitor having no potential difference across its plates. (For mathematical ease, put in a resistor to get an RC circuit and let resistance -> 0 at the end), close the switch and calculate the charging current. Note that q is not constant so setting work=qV is not correct, unless you work with differentials. In fact, a standard way to derive the basic circuit equations, LRC, is to start with a statement of conservation of energy in terms of circuit properties. With a little work, you'll be able to answer your own question.

Regards,
Reilly Atkinson

 

[maverick280857]

There is possibly another more fundamental explanation...based on the self energy of the capacitance. This comes from something I read in an electromagnetics text a few days ago and it appears to me that the self energy computations for the configuration being discussed should give us a clue as to where the (1/2) factor gets "eaten up".

Even though our general physics books discuss this as an outcome of a series resistance, it is probably better to think about self energy.

Here is a list of research papers which have addressed this problem more formally: http://www.smpstech.com/charge.htm (some of these may not be totally germane to the present discussion).

 

[ceptimus]

You might also think about the inherent paradox of attempting to charge a capacitor at a constant voltage. The whole definition of capacitance relates the voltage across it to the amount of charge present.