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.

How a capacitor works in a circuit

Time dependence of charging and discharging

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.

Unit of capacitance

The unit of capacitance is the Farad (F), named after Michael Faraday. The base units of the Farad are, $1 \; F = s^4 A^2 m^{-2} Kg^{-1}$. 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.

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, $\epsilon_r$, called the relative static permeability. $\epsilon_r = 1$ for air and can range from 1 for a vacuum (we actually define $\epsilon_r = 1$ for a vacuum) to over 12,000 for an inorganic compound called calcium copper titanate.

Mathematically, capacitance scales linearly with overlap area and $\epsilon_o$, 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 $\epsilon_o$ in the capacitance formula is called the dielectric constant, $\epsilon_o = 8.754 \times 10^{-12} \; F/m$, and it's there, like any proportionality constant in physics, to get the units right.

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.

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.

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

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:

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

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, $q_f$.

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

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 $q_f$, then subtract 1.

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 $V_f$. That's convenient because $e^0=1$:

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:

Practice problems

1. A 5,000 μF capacitor is wired in series with a 4,000 μF capacitor in a circuit. Calculate the total capacitance of the pair.

   Solution

   
   

   the total capacitance of capacitors wired in series is like the total resistance of resistors wired in parallel:

   $$C_{tot} = \left( \frac{1}{C_1} + \frac{1}{C_2} + \dots \right)^{-1}$$

   For this circuit, we get

   $$ \begin{align} C_{tot} &= \left( \frac{1}{5000 \times 10^{-6}} + \frac{1}{4000 \times 10^{-6}} \right)^{-1} \\[5pt] &= 450^{-1} \, F \\[5pt] &= 0.0022 \, F \\[5pt] &= 2.2 \, mF \end{align}$$




2. Two 2,000 nF capacitors are wired in parallel in a circuit. Calculate the total capacitance of the pair.

   Solution

   
   

   The total capacitance of capacitors wired in parallel is like the total resistance of n resistors wired in series:

   $$C_{tot} = C_1 + C_2 + \dots + C_n$$

   For this circuit, we get

   $$ \begin{align} C_{tot} &= 2000 \times 10^{-9} + 2000 \times 10^{-9} \\[5pt] &= 4000 \times 10^{-9} F \\[5pt] &= 4000 \, nF \end{align}$$




3. A 300Ω resistor, a 4000μ F capacitor, a switch and a 10V battery are wired in series in a circuit. Determine the initial and steady-state currents. Determine how long it will take the circuit to reach its steady state.

   Solution

   
   

   The total capacitance of capacitors wired in parallel is like the total resistance of n resistors wired in series:

   $$C_{tot} = C_1 + C_2 + \dots + C_n$$

   For this circuit, we get

   $$ \begin{align} C_{tot} &= 2000 \times 10^{-9} + 2000 \times 10^{-9} \\[5pt] &= 4000 \times 10^{-9} F \\[5pt] &= 4000 \, nF \end{align}$$




4. A capacitor with C = 4.2 × 10^-7 F is charged when electrons move from one plate, around the circuit to the other, achieving a potential (voltage) difference of 600 V. How many electrons were moved in order for the capacitor to achieve this charge?

   Solution

   
   

   The total capacitance of capacitors wired in parallel is like the total resistance of n resistors wired in series:

   $$C_{tot} = C_1 + C_2 + \dots + C_n$$

   For this circuit, we get

   $$ \begin{align} C_{tot} &= 2000 \times 10^{-9} + 2000 \times 10^{-9} \\[5pt] &= 4000 \times 10^{-9} F \\[5pt] &= 4000 \, nF \end{align}$$




5. Analyze this RC circuit. Calculate (a) the current in the loop after three time constants have elapsed, and (b) the energy stored in each capacitor once a steady state has been reached.

   

   Solution

   
   

   First, the total capacitance of this circuit is

   $$ \begin{align} C_{tot} &= \left( \frac{1}{20 \times 10^{-6}} + \frac{1}{30 \times 10^{-6}} \right)^{-1} \\[5pt] &= 12 \times 10^{-6} \, F = 12 \, \mu F \end{align}$$

   Now the RC time constant is

   $$RC = (50 \, \Omega)(12 \times 10^{-6} \, F) = 600 \, \mu s$$

   The