Springs, Capacitors and Devices 2
Maybe this will turn into a little series... this time capacitors!
Firstly, you might not know what a capacitor is, so here's how it was explained to me (the way I liked anyway): what happens if you have a circuit with a gap in it?
Well, the GCSE answer is that 'current does not flow'. But does this make sense from a fundamental perspective? What does a battery actually do? It maintains a potential difference across itself by producing an electric field. This electric field WILL cause electrons to move in a wire. FACT. So electrons will flow away from the negative side and towards the positive side precisely because opposites attract. But we know that current doesn't appear to flow, so something must be off about this...
What happens is that the charge can't go anywhere. It does start to move, but it will build up at the 'gap' and will set up an electric field that opposes or cancels the battery's. (Or in force terms, the battery applies some force to the electrons, and the charge built up repels them by the same amount). What a capacitor is, is a more efficient way of storing that built up charge. The capacitance is typically called C (for obvious reasons) and is defined by the equation:
Q = CV ; C = Q/V
Read as "Capacitance is the charge stored on an object (not yet specified) per unit voltage applied across it".
The typical capacitor is a pair of parallel plates. One can show, either by a horrific integral, or by Gauss' theorem, that the electric field between two plates is uniform.
E = Q/Ae ; Sorry! This is normally 'sigma over epsilon'. I can't write either of those easily, so PLEASE try to bear with me and please Google this so that you can see the proper equation. I'd try HyperPhysics.
Brief edit: A friend pointed out that E = sigma over epsilon is in fact, the equation defining Young's modulus. Just a nice little coincidence, and although the deeper connection to elastic materials like springs is not.
(Note: this is really a digression, but I want to cover the construction of a capacitor in brief detail, because everyone knows what a spring is, so you also need an idea of what a capacitor is to contrast them properly.)
For a uniform electric field we have V = Ed, where d is the separation across the plates. We have plates of area A and separation d. The voltage is V, and the charge Q, therefore:
C = Q/V = Q/Ed = Ae/d = e(A/d)
I rewrote the last equation just to point out that the only relevant thing about a parallel plate capacitor is the ratio of the area to the thickness. If you want a really good capacitor, then you need massive plates that are really close together. One problem with that: if they get too close, the air gap will be able to ionise and will start to conduct: now you have a circuit, or in the words of my Physics teacher, "you've just built a really expensive wire".
This gives us the units for epsilon as well. C has dimensions (charge)/(voltage) = (charge)/(work)/(charge) = (charge)^2/(work). We call that a farad and a typical capacitance is a micro or a Pico farad. Epsilon therefore has units of farads per metre. Epsilon, if you don't know, is the electric constant. It's the constant that pops up in all the electric equations, it's related to Coulomb's constant by the equation k = 1/(4pi e). It's really not that important at the moment, we haven't even gotten to capacitors in series yet!
So... I think that wraps up the construction of a capacitor for now. This lets us calculate the capacitance in series and in parallel just like for springs:
In series: you are going to have two capacitors of width d and d'. As you bring them together, nothing changes and when they're next to each other the capacitance had better still not change. So we want the new capacitance to be given by:
C(new) = Ae/(d+d') (remember, the area of the plates is the same, just the separation is increased.)
Fiddle around a bit, and you can show that this is the same as:
1/C = 1/C_1 + 1/C_2 ; This can be shown in the same way as for springs, because it's the same equation.
Now parallel: much easier, you haven't changed the separation, you've just added the areas together. This gives us:
C(new) = (A+A')e/(d) = C_1 + C_2
Hmm... that's the same as for springs in parallel. Again, not surprising as it's the same equation.
What else could we find out? The energy stored in a capacitor? Well, you are building up a charge and an associated electric field so there's some energy in there. What is it?
V = dW/dQ ; W = int [ V dQ]
we know V = Q/C so that W = int[Q/C dQ]
W = 1/2 Q^2/C = 1/2 C V^2 = 1/2 QV
For springs we had W = 1/2 k x^2 and W = 1/2 Fx ... this is starting to feel VERY familiar.
What about the energy density in a parallel plate capacitor? That's an interesting quantity if you're doing certain things. By this we mean 'energy divided by volume'
I.e. P = 1/2CV^2/ (Ad) for the parallel plate capacitor. This simplifies (try it) to:
P = E^2 e/2
Two things to note: firstly this equation is way more general than its derivation. Physics has a funny habit of doing that sometimes - you can derive conservation of angular momentum for planetary orbits from Newton's laws. It's just that that doesn't explain why it's also true for ballerinas, roundabouts, and well... everything.
Secondly is the symbol P. It's appropriate, because it looks like a rho (for density) and it just so happens that energy density has the same unit as pressure: they're both measured in Pascals.
So, springs and capacitors are the same thing. We get it. You put them in series and parallel and the same stuff happens, you stretch a spring you store energy, you charge a capacitor you store energy. You release a spring and it oscillates, you release a capacitor and...
It doesn't oscillate.
This post is getting pretty damn long, but this is worth saying: maybe it does oscillate a little, especially if it's on its own in a circuit (causing some build up of charge somewhere that repels and reverses the current). But it's certainly not the primary effect. The main effect is the capacitor discharges. Quite gently, as things go (it's exponential, so not gently, but... you know what I mean... it doesn't oscillate).
The current is caused by the potential difference across the capacitor:
V = IR, if there is some load resistance R attached (which you want, obviously, you can't discharge a capacitor without anything attached.)
But it's a capacitor and so V = Q/C = IR
I = (1/RC) Q
The value RC has units (volt/amp) * (coulomb / volt) = seconds. RC is called the time constant for the circuit because it's the most natural quantity that is a time to associate with it.
The equation I = Q/RC is a differential equation because I = dQ/dt. It's the easiest kind (1st order separable) so we can solve it:
Q = Q_0 e^-(t/RC)
OR, if we use RC as out time scale so that t' is a dimensionless number telling us how many time constants it's been: Q = Q_0 e^-t'.
I.e. exponential decay (loosely speaking the characteristics are: rapid decrease at start, slow decrease later, constant half life, charge goes down by a factor of e (2.72) every time constant, i.e. it goes to about 37% of its starting value).
Let's summarise and be done with this REALLY long post:
Parallel plate capacitor: C = eA/d
Series capacitor (all): 1/C = sum [1/C_i]
Parallel capacitor (all): C = sum [C_i]
Energy density due to electric field: P = e/2 E^2 (E^2 dependence is the important thing.)
Energy stored: W = 1/2 CV^2 = 1/2 QV
Differential equation : dQ/dt = I = Q/RC
Solution: Q(t) = Q(0)e^-t/RC = Q(0)e^-t'
I'll just let that lie like that.
