Work done by load in charging a capacitor

mfb

And this is reasonable, as the energy released EM radiation is really small, unless you add some fancy oscillators to the setup, charge it with another capacitor and superconducting wires or do something else unusual I don't think of at the moment.

Resistance is always there, too, unless you use superconductors everywhere.

Right.

truesearch

Would any one agree that the answer to the original post:
"the first answer implies, i think, that energy is required to move charges against the already present charges on the capacitor plate. is this correct?"
is incorrect and a good answer would be :
The energy lost is due to
a) Heat produced in any resistance
b) electromagnetic radiation from the connecting wires
c) sparks (if there are any)

sophiecentaur

It's certainly always there but it can hardly be thought of as the major loss in most 'real' circumstances. In many ways, it can be looked upon as a bit of a distraction for a student who is probably struggling with the basics of circuits and Maths. It appears as a resistance in series with the Ohmic resistances in the circuit and can be included in the little rectangle that's always drawn in the 'charging a capacitor' circuit.
'Sensible' explanations don't usually bring in yet another can of worms for a poor student to deal with. A level doesn't actually deal with launching em waves into space. What level do you teach?

sambristol

folks check out this

http://www.physics.princeton.edu/~mc...es/twocaps.pdf

Regards

Sam

sweet springs

Hello.

The More charges gather to the plates, The More work we need to mount additional charge against repulsion forces. Thus the whole work are integration of V dQ = Q/C dQ. Thus 1/2 appears from formula of calculus, as well as it appears in Mechanics the distance is expressed 1/2 a t^2 by time t and acceleration a or in Mathematics of area of triangle.

Regards.

sophiecentaur

That's all fine and accurate but the power supply has delivered QV of energy - twice as much as that. That's where people find the difficulty.

truesearch

I teach A level physics and electronics.
Sensible explanations do not avoid covering physics in detail.
A level does deal with 'launching em waves into space' Students need to know about the whole range of electromagnetic waves, how they are produced and how they are transmitted. We need to cover electromagnetic induction, fields due to current carrying conductors etc. It is easy to demonstrate transmission of electromagnetic waves using 2 coils, a signal generater and an oscilloscope.
The most basic work on capacitors requires an understanding of the energy stored and why it is 0.5QV. When a student asks (and they do ask!!) what happens when resistance is zero (they have heard of superconductors at A level) they deserve an explanation.
The explanation is easy....it has been covered in this forum before.

mfb

The question is "how detailed?"
Would you skip Newtonian mechanics, as GR is more precise? ;)
Or skip an introduction to classical EM waves, as quantum electrodynamics is more precise?
Of course not - both are beyond the scope of your classes.

While it is useful to know that there is something more behind this, it is a good idea to use the easiest explanation which gives good results. For an object falling in the lab, this is Newtonian gravity. For charging a capacitor, this is energy losses in a resistor. Most regular power supplys have something similar to an internal resistance, even if you don't use an external one.
It might be interesting to consider the setup with two capacitors and superconductors, but it is certainly not the main point the students should learn about charging capacitors.

sweet springs

Hi.

Why do you have such an idea? I suppose you believe that voltage between plates is constant during charging immediately after the moment the condenser is connected to battery. When battery is connected to the condenser, V=0 because no charge at the plates mean no voltage between. V would increase as charge is accumulated on the plates.

For example tall glass of water and short glass of water is connected by tube so that water can move between. Is water level equal at the moment of the connection?

Regards.

sophiecentaur

Of course not. Your connecting pipe requires a pressure drop along it in order to allow a flow of water. The Maths correspond to an RC circuit and not just a pair of Capacitors. The water transfer takes TIME, during which, energy is dissipated.
If we are discussing a real situation in which there is some series resistance and a source of emf, then, of course, there will be a finite time during which the capacitor becomes charged. The source of emf will have been producing (by definition) V volts all the time. The PD across the capacitor will have started at zero and eventually reached V. During this time, current is flowing through the resistive element (which always has a real, radiative component). QV is the energy supplied by the source. QV/2 is the energy stored in the capacitor. The rest has been dissipated.
If you want to do the analysis more thoroughly then you could introduce the Inductive properties of any connecting wires. If the L is high enough, then the reactance is high enough to treat the system as a damped oscillator then you will get a different path towards the equilibrium condition but the same end result applies.

sophiecentaur

The explanation of how EM waves are launched into space is far from easy. Why do you think it is not dealt with (i.e. analysed) at A level? Electromagnetic induction is dealt with very superficially (appropriately so). The fact that em waves can be radiated from objects is fairly easy to take on board but how many 'informed' adults ever make the connection between the good old H atom radiates (involving photons) and the way a wire dipole radiates 10MHZ signals? Would you also recommend using QM to discuss the way a capacitor charges?
I would also take issue when you state that they have "heard of superconductivity", implying that they have not confused it with 'very high conductivity'. Students have a very limited scientific life experience (not helped by the TV these days) and they very easily confuse Science with Science Fiction. Down - to - earth explanations get my vote every time.

As a teacher, I would have assumed that you would take into account the limitations of your students and that you'd avoid confusing their superficial fascination with actual understanding. Or perhaps you have a particularly bright set of students, in which case you are very lucky.

sweet springs

Hi. Now I think I get your point.

In the example of connecting pipe, you will see see-saw of water levels in both the glasses and it continues eternally if the liquid is perfect fluid.

Similarly the increasing electric field between plates of capacitor by rushing charges cause magnetic field. Disappear of magnetic field cause excess current to the capacitor. In the case battery is also charged capacitor of the same type, see-saw of charges begin. If all the dynamic parts of energy dissipate, the remaining static part of energy is half the initial one.

Regards.

sophiecentaur

That's it.
The argument continues as to the relative amounts of dissipation by the various mechanisms.

DaleSpam

Actually, the energy dissipated in the resistance for a charging series RC circuit is independent of the actual value of the resistance. For a series RC circuit with initial voltage 0 across the capacitor and with a constant supply voltage v_s after t=0 we have the differential equation:
[tex]C \frac{dv_c}{dt}+\frac{v_c-v_s}{R}=0[/tex]
Which has the solution:
[tex]v_c=v_s(1-e^{-\frac{t}{RC}})[/tex]

So the energy dissipated in the resistor is given by:
[tex]\int_0^{\infty} i_r v_r \, dt = \int_0^{\infty} \frac{v_c-v_s}{R} (v_c-v_s) \, dt = C \frac{v_s^2}{2}[/tex]
which is not a function of R, and is also exactly equal to the "missing" energy.

As to whether or not resistance or radiation is the dominant source, if you are using circuit theory or doing circuit analysis then one of the basic assumptions/approximations that you are making is that the radiation is 0. Even making the assumption of no radiation, you can still show that energy is always conserved, so it is not necessary to invoke radiation to explain the missing energy.

sophiecentaur

You are so right.
And it amazes me that people don't believe the two values of energy involved - QV and QV/2 are enough to prove the point. You shouldn't have to go as far as you have done in order to prove it.

truesearch

VrxIr tells you how much energy (power) is dissipated by the wire. It does not tell you what form of energy is dissipated.
You could assume there was no Joule heating just as easily as you could assume there is no radiation. You are quite right to recognise that circuit analysis does not reveal the forms of energy involved.
It looks like you have taken the path that proves there is no electromagnetic radiation.
Just for the moment I like the assumption that there is no Joule heating.
Have you considered 'radiation resistance' in your analysis?

sophiecentaur

I mentioned Radiation Resistance way back in the thread. There is no detectable difference as far as the source is concerned although a quick frequency sweep could allow R(ohmic) and R(radiation) to be identified.