 Passing current through empty space

Capacitors are very interesting and very useful circuit elements. They can store charge, discharge it rapidly (or more slowly if used in combination with resistors), and they can essentially block direct current. They can be used in timing circuits, filtering circuits ... all kinds of things. Rapid chargers in electric cars employ capacitors to store charge rapidly and introduce it to the batteries more slowly later.

Anatomy of a capacitor

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. How a capacitor works in a circuit  An object is electrically polarized when there is a significant charge imbalance between two of its sides or ends. The plates of a capacitor become polarized and depolarized in use. When current flows to an un-charged capacitor, it is initially large. But as the capacitor becomes more polarized, it becomes more difficult to polarize further, so the rate of polarization decreases. The graph shows the basic shape of a current vs. time graph for a capacitor.

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 t = 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. Unit of capacitance

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 1 F 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

Capacitance depends on geometry and insulation

The amount of capacitance (capacity to hold charge) of such a device is dependent on three things:

1. The area of overlap of the two parallel plates, A
2. The distance between the plates, d
3. The nature of the material in between the plates, which cannot be a conductor. 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

• small inter-plate distances,
• large areas of overlap between parallel plates, and
• a separating material with a very high permittivity value*.

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.

A better kind of insulation for a capacitor: dielectrics

We don't want the material (air, liquid or solid) between capacitor plates to conduct electricity so it needs to be an insulator, but that doesn't mean we can't choose an insulating material that helps us make a better capacitor. Some materials are dielectrics. They are insulators, yet when we apply an electric field to them, as we would between the plates of a capacitor, their polar subunits tend to re-orient along the direction of the field.

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, and not polarized, like this. If we apply an electric field to such a dielectric material, the polar constituent molecules can line up and serve to help polarize the two capacitor plates. Many of the materials in the table above, including ceramics, plastic films and minerals like mica are good dielectrics.

Capacitance in RC-circuits

Parts of these derivations involve a little calculus. You can still use the result if you can't follow the calculus.

We're going to develop a few equations to understand RC circuits. The simplest RC circuit consists of a resistor, capacitor and battery in series like the one below.

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):

$$C = \frac{q}{V}$$

Notice that we could also solve for voltage and write V = q/C.

Deriving the charging equations

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.

$$R \frac{dq}{dt} = V = \frac{q}{C}$$

To begin to separate this equation, isolate dq/dt on the left by dividing by R:

$$R \frac{dq}{dt} = \frac{1}{R} \left( V - \frac{q}{C} \right)$$

Now separate: divide by V-q/C on the left and multiply both sides by the differential dt.

$$\frac{dq}{V - \frac{q}{C}} = \frac{dt}{R}$$

Now we can multiply V by C/C to get a single fraction in the left-side denominator,

$$\frac{dq}{\frac{CV - q}{C}} = \frac{dt}{R}$$

... 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:

$$\frac{dq}{CV - q} = \frac{dt}{RC}$$

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.

$$\int_0^{q(t)} \frac{dq}{CV - q} = \frac{1}{RC} \int_0^t \, dt$$

That's a simple integral to do, and the right-side limits are easy to evaluate:

$$-ln|CV - q| \bigg|_0^{q(t)} = \frac{t}{RC}$$

Now evaluate the left-side limits.

$$-ln|CV - q(t)| + ln|CV| = \frac{t}{RC}$$

It will be easier to go ahead if we multiply everything by -1:

$$ln|CV - q(t)| - ln|CV| = -\frac{t}{RC}$$

Use the division law of logs to rearrange:

$$ln \left| \frac{CV - q(t)}{CV} \right| = -\frac{t}{RC}$$

... and exponentiate both sides to get closer to a nice exponential form.

$$\frac{CV - q(t)}{CV} = e^{-\frac{t}{RC}}$$

Now let's begin solving for q(t) by separating (CV-q(t))/CV into two fractions, CV/CV and -q(t)/CV:

$$1 - \frac{q(t)}{CV} = e^{-\frac{t}{RC}}$$

Move the 1 to the right and switch the signs again:

$$\frac{q(t)}{CV} = 1 - e^{-\frac{t}{RC}}$$

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

$$q(t) = q_f \left( 1 - e{-\frac{t}{RC}} \right)$$

Time-dependent voltage charging equation

Now one more thing. If Vc is the voltage across the capacitor, then we can use the relationships

$$V_c(t) = \frac{q(t)}{C}$$

and

$$\frac{q_f}{C} = V_f = V_{battery}$$

to divide both sides by the capacitance to get a similar equation for the time-dependent voltage.

$$V(t) = V_f \left( 1 - e^{-\frac{t}{RC}} \right)$$

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.

time-dependent charge

$$q(t) = q_f \left( 1 - e^{\frac{-t}{RC}} \right)$$

time-dependent potential

$$V(t) = V_f \left( 1 - e^{\frac{-t}{RC}} \right)$$

Discharging a capacitor

Now go back to the RC circuit diagram above and let's charge the capacitor and set the switch to position b, so that current flows through the resistor to discharge the capacitor. In that case, there is no more battery, and the loop rule says that the potential across the capacitor has to equal that across the resistor, and that they have to sum to zero, so Substituting dq/dt for the current, we get:

$$R \frac{dq}{dt} = -\frac{q}{C}$$

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:

$$\int \frac{dq}{dt} = -\int \frac{dt}{RC}$$

The solution is straightforward. Here A is the constant of integration

$$ln|q| = -\frac{t}{RC} + A$$

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:

$$q(t) = A e^{-\frac{t}{RC}}$$

The boundary condition is q(0) = qf. That is, at time t = 0, the capacitor is fully-charged. So A = qf.

$$q(t) = q_f e^{-\frac{t}{RC}}$$

As we did for the charging equations above, we can divide both sides by the capacitance, C, to get the time-dependent voltage equation:

$$V(t) = V_f e^{-\frac{t}{RC}}$$

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. time-dependent charge

$$q(t) = q_f e^{\frac{-t}{RC}}$$

time-dependent potential

$$V(t) = V_f e^{\frac{-t}{RC}}$$

Example 1

An RC series circuit driven by a 12.0 V battery, with R = 1.5 MΩ and C = 1.70 μF. Calculate the time constant, find the maximum charge that will gather on the capacitor, and calculate the time it will take to build up a 10 μC charge.

Solution: The time constant of an RC circuit is just the product of the resistance, R, in Ohms and the capacitance, C, in Farads.

$$RC = (1.5 \times 10^6 \, \Omega)(1.7 \times 10^{-6} \, F) = 2.55 \; s$$

The charge on a capacitor as a function of time (from the derivation above) is: The product of the capacitance (C) and voltage of the power supply (V) is the final charge, or the asymptotic limit of the charge on the capacitor – the maximum amount of charge it can carry, qf.

$$q(t) = q_f \left( 1 - e^{-\frac{t}{RC}} \right)$$

To find the time it would take to gain a charge of 0.2 μC, we rearrange to solve for time. First divide both sides by qf, then subtract 1.

Here I've also multiplied both sides by -1 for convenience:

$$1 - \frac{q(t)}{q_f} = e^{-\frac{t}{RC}}$$

Taking the natural log of both sides gives:

$$ln \left( 1 - \frac{q(t)}{q_f} \right) = \frac{-t}{RC}$$

And finally multiplying both sides by RC = 2.55 s gives us the time. I'm not placing absolute value bars on the log function because qf should always be greater than any other charge we could obtain on the capacitor.

$$t = -RC \cdot ln \left( 1 - \ln{q(t)}{q_f} \right)$$

The result is

$$t = -2.55 \, s\cdot ln\left( 1 - \frac{10 \times 10^{-6} \; C}{2.04 \times 10^{-5} \; C} \right)$$

and   $\bf t = 1.72 \; s$.

Example 2

A capacitor with an initial potential difference of 100 V is discharged through a resistor when a switch between them is closed at t = 0. At t = 10.0 s, the potential difference across the capacitor is 1.00 V. (a) Calculate the time constant of the circuit. (b) Calculate the potential difference across the capacitor at t = 15.0 s.

Solution: The discharge potential as a function of time is given by the formula we derived above,

$$V(t) = V_f e^{\frac{t}{RC}}$$

We have V(0), so we can solve for Vf. That's convenient because e0 = 1: Now we can rewrite V(t) as

$$V(t) = 100 e^{-\frac{t}{RC}}$$

Now we have another data point, V(10) = 1.0, that we can use to find RC. We'll solve for RC first by dividing by 100 on both sides:

$$\frac{V(t)}{100} = e^{-\frac{t}{RC}}$$

Now take the natural log on both sides:

$$-ln \left( \frac{V(t)}{100} \right) = \frac{t}{RC}$$

Finally, solve for RC:

$$RC = \frac{-t}{ln \left( \frac{V(t)}{100} \right)}$$

Plugging in t = 10 and V(10) = 1.0 V, we get the RC time constant for this circuit:

$$RC = \frac{-10}{ln \left( \frac{1}{100} \right)} = 2.17 \; s$$

Now the final discharging potential equation is

$$V(t) = 100 e^{\frac{-t}{2.17}}$$

Finally, at time t = 15s, the potential across the capacitor is

$$V(15) = 100 e^{\frac{-15}{2.17}} = 0.1 \; V$$   xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.