A capacitor is really pretty simple. It's two parallel plates, separated by air or some other insulator. The plates must be pretty close together, often just the with of a thin plastic film.. Each plate is connected to a conductor which connects into a circuit.
An object is electrically polarized when there is a significant charge imbalance between two of its sides or ends.
Notice that electrons don't really pass through the gap between the capacitor plates. The presence of a new negative charge on one plate (left plate in the images above) pushes a negative charge away from the opposite plate through electrostatic repulsion. The electrostatic force is a force that acts at a distance (without touching), like gravity.
The times in this figure correspond roughly to the 1,2,3,4 diagram above.
Here's what a capacitor might look like in practice. They come in a wide variety of physical sizes, but they're usually cylindrical or disk-shaped.
The unit of capacitance is the Farad (F), named after Michael Faraday. The base units of the Farad are, 1 F = s4·A2·m-2·Kg-1. That's not often too helpful. The most useful conversion we can make (and there are a lot of them) is that 1 F = 1 C/V; One Farad is one coulomb of charge per volt of potential.
In practice, one Farad (1F) is a very large amount of capacitance. Charge a 1F capacitor up and touch it and you'll get the shock of your life (maybe your last). In electronics applications, units of microfarads (1 μF = 10-6 F), nanofarads (1 nF = 10-9 F) or picofarads (1 pF = 10-12 F) are much more common. Pictured above is a 68 μF capacitor that can be held at a potential of up to 100 V before the insulator between the plates breaks down and allows current to flow.
The unit of capacitance is the Farad.
1 F = 1 C/V
The two geometric properties, A & d, aren't difficult to figure out. The quality of insulation between the plates is ranked with a number, εr, called the relative static permeability. εr = 1 for air and can range from 1 for a vacuum (we actually define εr = 1 for a vacuum) to over 12,000 for an inorganic compound called calcium copper titanate.
Mathematically, capacitance scales linearly with overlap area and εr, and inversely with the distance between plates:
It's easy to see that as we increase either the area of overlap of the plates or the quality of the insulation material, and as we decrease the the distance between plates, we increase the capacitance of the device. The constant εo in the capacitance formula is called the dielectric constant, εo = 8.754 x 10-12 F/m, and it's there, like any proportionality constant in physics, to get the units right.
Here are some relative permittivity values for a few substances.
In capacitors, we'd like to have
We get small distances by using very thin insulating materials. Hemp fiber has recently become a good choice because its fibers are thin, cheap and have a high εr value. In order to have more overlap area, many capacitors employ plate pairs that are rolled up into a cylinder, preserving the constant distance between them, but dramatically enlarging the overlap area.
*Note that the word "permittivity" here is the opposite of what we really mean. High-permittivity substances really do not "permit" the flow of current very well. It's an old term we're kind of stuck with.
Here is a schematic diagram of a dielectric material, made of [+ -] dipoles that is not in an electric field. The dipole constituents are oriented pretty much randomly.
If we apply an electric field, the polar constituent molecules can line up and serve to help polarize the two plates.
Parts of these derivations involve a little calculus. You can still use the result if you can't follow the calculus.
I've dressed that circuit up just a bit to include a volt meter across the capacitor so we can look at the time-dependent voltage across the capacitor (the potential will change as the capacitor charges), and a fancy switch, S.
When the switch is in position a, the battery charges the capacitor, with its current limited by the resistance, R. When it's in position b, the charge on the capacitor can leak away as current through the resistor until the two plates are once again unpolarized.
For a two-plate capacitor like the one above, capacitance is defined as charge divided by the potential difference between the plates, where the charge, q, is the charge on either plate (+q on one, -q on the other):
Notice that we could also solve for voltage and write V = q/C.
The loop rule of circuits says that the voltage increase, V, is equal to the voltage drop across the resistor, IR plus the voltage drop across the capacitor, q/C. We write the current in its derivative form: the change in charge divided by the change in time.
Here's our differential equation. It's separable.
To begin to separate this equation, isolate dq/dt on the left by dividing by R:
Now separate: divide by V-q/C on the left and multiply both sides by the differential dt.
Now we can multiply V by C/C to get a single fraction in the left-side denominator,
... and recognize that division is just multiplication by the reciprocal. That leaves a C multiplying the dq on the left. Divide it to the right side:
Now we can integrate, and we'll do it from time t = 0 (at which point we assume that the charge on the capacitor is zero) to time t, at which point the time-dependent charge is class="center". We're summing up charge here.
That's a simple integral to do, and the right side limits are easy to evaluate:
Now evaluate the left-side limits.
It will be easier to go ahead if we multiply everything by -1:
Use the division law of logs to rearrange:
... and exponentiate both sides to get closer to a nice exponential form.
Now let's begin solving for q(t) by separating (CV-q(t))/CV into two fractions, CV/CV and -q(t)/CV:
Move the 1 to the right and switch the signs again:
And finally multiply by CV. Notice that CV is just the charge, but it's the final charge because V is the potential of the battery. We'll call that qf.
Our final equation for the time-dependent charge on a capacitor is
Now one more thing. If Vc is the voltage across the capacitor, then we can use the relationships
to divide both sides by the capacitance to get a similar equation for the time-dependent voltage.
Finally, here's a graph of f(t) = 1 - e-t, Which is either one of our equations with V, R, C = 1, just to get an idea of the shape of the graph.
You can see that the capacitor charges rapidly at first, but more slowly as it becomes more fully polarized. Changing R, C and V would'nt change that essential feature, only how long the charging takes.
Substituting dq/dt for the current, we get:
Now we can separate variables (divide by q on both sides and multiply by dt) to get a form of the differential equation that we can integrate:
The solution is straightforward. Here A is the constant of integration
As usual in such equations, we can make that constant multiplicative after we exponentiate both sides to get q(t) out of the ln( ) expression:
The boundary condition is q(0) = qf. That is, at time t = 0, the capacitor is fully-charged. So A = qf.
As we did for the charging equations above, we can divide both sides by the capacitance, C, to get the time-dependent voltage equation:
Here is a plot of the function q(t) = e-t, and you can see that it's just the mirror image of the charging curve. The parameters V, R and C would modify the shape a bit, but it's still an exponential decay. At the beginning, the driving force for the discharge current comes from the high polarization of the capacitor plates, but as time goes on, that force diminishes.
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