Capacitors

It is clear that we can store electric charge on the surface of a conductor. However, electric fields will be generated immediately above this surface. Now, the conductor can only successfully store charge if it is electrically insulated from its surroundings. Of course, air is a very good electrical insulator. Unfortunately, air ceases to be an insulator when the electric field-strength through it exceeds some critical value which is about $E_{\rm crit} \sim 10^6$ volts per meter. This phenomenon, which is known as breakdown, is associated with the formation of sparks. The most well-known example of the breakdown of air is during a lightning strike. Thus, a good charge-storing device is one that holds a relatively large amount of charge, but only generates relatively small external electric fields (so as to avoid breakdown). Such a device is called a capacitor.

Consider two thin, parallel, conducting plates of cross-sectional area $A$ that are separated by a small distance $d$ (i.e., $d\ll \!\sqrt{A}$). Suppose that each plate carries an equal and opposite charge $\pm Q$ (where $Q>0$). We expect this charge to spread evenly over the plates to give an effective sheet charge density $\pm\sigma = Q/A$ on each plate. Suppose that the upper plate carries a positive charge and that the lower carries a negative charge. According to Equation (2.130), the field generated by the upper plate is normal to the plate and of magnitude

$$E_{\rm upper} =\left\{ \begin{array}{lcl} +\sigma /(2 \,\epsi... ...psilon_0)& \mbox{\hspace{2cm}}&\mbox{below} \end{array}\right..$$ (2.136)

Likewise, the field generated by the lower plate is

$$E_{\rm lower} =\left\{ \begin{array}{lcl} -\sigma /(2\, \epsi... ...psilon_0)& \mbox{\hspace{2cm}}&\mbox{below} \end{array}\right..$$ (2.137)

Note that we are neglecting any leakage of the field at the edges of the plates. This is reasonable provided that the plates are relatively closely spaced. The total electric field is the sum of the two fields generated by the upper and lower plates. Thus, the net field is normal to the plates, and of magnitude

$$E_\perp =\left\{ \begin{array}{lcl} \sigma /\epsilon_0&\mbox{... ...ex] 0& \mbox{\hspace{2cm}}&\mbox{otherwise} \end{array}\right..$$ (2.138)

See Figure 2.7. Because the electric field between the plates is uniform, the potential difference between the plates is simply

$$V = E_\perp\,d = \frac{\sigma \,d}{\epsilon_0}.$$ (2.139)

[See Equation (2.17).]

Figure 2.7: The electric field of a parallel plate capacitor.

It is conventional to measure the capacity of a conductor, or set of conductors, to store electric charge, but generate small external electric fields, in terms of a parameter called capacitance. This parameter is usually denoted $C$. The capacitance of a charge storing device is simply the ratio of the charge stored to the potential difference generated by this charge:

$$C = \frac{Q}{V}.$$ (2.140)

Clearly, a good charge storing device has a high capacitance. Incidentally, capacitance is measured in farads (F), which are equivalent to coulombs per volt. This is a rather unwieldy unit, because capacitors in electrical circuits typically have capacitances that are only about one millionth of a farad.

For a parallel plate capacitor, we have

$$C = \frac{\sigma\,A}{V} = \frac{\epsilon_0\,A}{d}.$$ (2.141)

Note that the capacitance only depends on geometric quantities, such as the area and spacing of the plates. This is a consequence of the superposability of electric fields. If we double the charge on a set of conductors then we double the electric fields generated around them, and we, therefore, double the potential difference between the conductors. Thus, the potential difference between the conductors is always directly proportional to the charge on the conductors. Moreover, the constant of proportionality (the inverse of the capacitance) can only depend on geometry.

Suppose that the charge $\pm Q$ on each plate of a parallel plate capacitor is built up gradually by transferring small amounts of charge from one plate to another. If the instantaneous charge on the plates is $\pm q$, and an infinitesimal amount of positive charge $dq$ is transferred from the negatively charged to the positively charge plate, then the work done is $dW= V\,dq = q\,dq/C$, where $V$ is the instantaneous voltage difference between the plates. (See Section 2.1.5.) Note that the voltage difference is such that it opposes any increase in the charge on either plate. The total work done in charging the capacitor is

$$W =\frac{1}{C} \int_0^Q q \,dq = \frac{Q^2}{2\,C} = \frac{1}{2}\,C\,V^2,$$ (2.142)

where use has been made of Equation (2.140). The energy stored in the capacitor is the same as the work required to charge up the capacitor. Thus, the stored energy is

$$W= \frac{1}{2}\, C \,V^2.$$ (2.143)

This is a general result that holds for all types of capacitor.

The energy of a charged parallel plate capacitor is actually stored in the electric field generated between the plates. This field is of approximately constant magnitude $E_\perp = V/d$, and occupies a region of volume $A\,d$. Thus, given the energy density of an electric field, $U = (\epsilon_0/2)\,E^{2}$ [see Equation (2.84)], the energy stored in the electric field is

$$W = \frac{\epsilon_0}{2} \frac{V^2}{d^2}\, A \,d= \frac{1}{2}\, C\, V^2,$$ (2.144)

where use has been made of Equation (2.141). Note that Equations (2.142) and (2.144) agree with one another. The fact that the energy of a capacitor is stored in its electric field is also a general result.

The idea, that we discussed in the previous section, that an electric field exerts a negative pressure $(\epsilon_0/2)\,E_\perp^{\,2}$ on conductors immediately suggests that the two plates in a parallel plate capacitor attract one another with a mutual force

$$F = \frac{\epsilon_0}{2}\, E_\perp^{~2}\,A= \frac{1}{2} \frac{C\, V^2}{d}.$$ (2.145)

It is not actually necessary to have two oppositely charged conductors in order to make a capacitor. Consider an isolated conducting sphere of radius $a$, centered on the origin, that carries an electric charge $Q$. The spherically symmetric, radial electric field generated outside the sphere is given by

$$E_r(r>a) = \frac{Q}{4\pi\,\epsilon_0 \,r^2},$$ (2.146)

and the associated electric potential is

$$\phi(r>a) = \frac{Q}{4\pi\,\epsilon_0 \,r}.$$ (2.147)

(See Section 2.1.7.) Here, $r$ is a spherical polar coordinate. (See Section A.23.) It follows that the potential difference between the sphere and infinity—or, more realistically, some large, relatively distant reservoir of charge such as the Earth—is

$$V = \frac{Q}{4\pi\,\epsilon_0 \,a}.$$ (2.148)

Thus, the capacitance of the sphere is

$$C = \frac{Q}{V} = 4\pi\,\epsilon_0\, a.$$ (2.149)

The energy of a spherical capacitor when it carries a charge $Q$ is again given by $(1/2)\, C\,V^2$. It can easily be demonstrated that this is equivalent to the energy contained in the electric field surrounding the capacitor.

Suppose that we have two spheres of radii $a$ and $b$, respectively, that are connected by a long electric wire. See Figure 2.8. The wire allows electric charge to move back and forth between the spheres until they reach the same potential (with respect to infinity). Let $Q_a$ be the charge on the first sphere, and $Q_b$ the charge on the second sphere. Of course, the total charge $Q_0= Q_a +Q_b$ carried by the two spheres is a conserved quantity. It follows from Equation (2.148) that if the spheres are at the same potential then

$$\frac{Q_a}{Q_0} = \frac{a}{a+b},$$ (2.150)

$$\frac{Q_b}{Q_0} = \frac{b}{a+b}.$$ (2.151)

Note that if one sphere is much smaller than the other one—for instance, if $b\ll a$—then the large sphere grabs most of the charge; that is,

$$\frac{Q_a}{Q_b} \simeq \frac{a}{b} \gg 1.$$ (2.152)

The ratio of the electric fields generated just above the surfaces of the two spheres follows from Equations (2.146) and (2.152):

$$\frac{E_b}{E_a}\simeq \frac{a}{b}.$$ (2.153)

Note that if $b\ll a$ then the field just above the smaller sphere is far larger than that above the larger sphere. Equation (2.153) is a simple example of a far more general rule; namely, the electric field directly above some point on the surface of a conductor is inversely proportional to the local radius of curvature of the surface.

Figure 2.8: Two conducting spheres connected by a wire.

It is clear that if we wish to store significant amounts of charge on a conductor then the surface of the conductor must be made as smooth as possible. Any sharp spikes on the surface will inevitably have comparatively small radii of curvature. Intense local electric fields are thus generated around such spikes. These fields can easily exceed the critical field for the breakdown of air, leading to sparking and the eventual loss of the charge on the conductor. Sparking can also be very destructive, because the associated electric currents flow through very localized regions, giving rise to intense ohmic heating.

As a final example, consider two co-axial conducting cylinders of radii $a$ and $b$, where $a<b$. Suppose that the charge per unit length carried by the outer and inner cylinders is $+\lambda$ and $-\lambda$, respectively. We can safely assume that ${\bf E} = E_r(r)\,{\bf e}_r$, by symmetry (adopting standard cylindrical polar coordinates). (See Section A.23.) Let us apply Gauss's law (see Section 2.4) to a cylindrical surface of radius $r$, co-axial with the conductors, and of length $l$. For $a<r<b$, we find that

$$2\pi \,r\, l\,E_r(r) = \frac{\lambda\,l}{\epsilon_0},$$ (2.154)

so that

$$E_r = \frac{\lambda}{2\pi\,\epsilon_0\,r}$$ (2.155)

for $a<r<b$. It is fairly obvious that $E_r=0$ if $r$ is not in the range $a$ to $b$. The potential difference between the inner and outer cylinders is [see Equation (2.17)]

$$V = - \int_{\rm outer}^{\rm inner} {\bf E} \cdot d{\bf r} = \int_{\... ...} = \int_a^b E_r\,dr = \frac{\lambda}{2\pi\, \epsilon_0} \int_a^b \frac{dr}{r},$$ (2.156)

so

$$V = \frac{\lambda}{2\pi\, \epsilon_0} \ln\left(\frac{b}{a}\right).$$ (2.157)

Thus, the capacitance per unit length of the two cylinders is

$$C = \frac{\lambda}{V} = \frac{2\pi\,\epsilon_0}{\ln (b/a)}.$$ (2.158)