Most capacitors have a nonconducting material, or dielectric, between their conducting plates. A common type of capacitor uses long strips of metal foil for the plates, separated by strips of plastic sheet such as Mylar. A sandwich of these materials is rolled up, forming a unit that can provide a capacitance of several microfarads in a compact package (Fig. 1).
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| Fig.1: A common type of capacitor uses dielectric sheets to separate the conductors. |
Placing a solid dielectric between the plates of a capacitor serves three functions. First, it solves the mechanical problem of maintaining two large metal sheets at a very small separation without actual contact.
Second, using a dielectric increases the maximum possible potential difference between the capacitor plates. Any insulating material, when subjected to a sufficiently large electric field, experiences a partial ionization that permits conduction through it. This is called dielectric breakdown. Many dielectric materials can tolerate stronger electric fields without breakdown than can air. Thus using a dielectric allows a capacitor to sustain a higher potential difference and so store greater amounts of charge and energy.
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| Fig.2:Effect of a dielectric between the plates of a parallel-plate capacitor. (a) With a given charge, the potential difference is $V_0$ (b) With the same charge but with a dielectric between the plates, the potential difference V is smaller than $V_0$. |
With the understanding of the behaviour of dielectrics in an external field developed in Section 2.10, let us see how the capacitance of a parallel plate capacitor is modified when a dielectric is present. As before, we have two large plates, each of area A, separated by a distance d. The charge on the plates is $\pm Q$, corresponding to the charge density $\pm \sigma$ (with $\sigma = Q/A$). When there is vacuum between the plates,
$E_0=\frac{\sigma}{\epsilon_0}$
and the potential difference $V_0$ is
$V_0=E_0d$
The capacitance $C_0$ in this case is
$C_0=\frac{Q}{V_0}=\epsilon_0 \frac{A}{d}$ (1)
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| Fig.3: Electric field lines with (a) vacuum between the plates and (b) dielectric between the plates. |
Consider next a dielectric inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field and, as explained in the previous section, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities $\epsilon_i$ and $-\epsilon_i$ (Fig.3). The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is $\pm (\sigma – \sigma_i)$. That is,
$E=\frac{\sigma-\sigma_i}{\epsilon_0}$ (2)
so that the potential difference across the plates is
$V=Ed=\left(\frac{\sigma-\sigma_i}{\epsilon_0}\right)d$ (3)
For linear dielectrics, we expect $\sigma_i$ to be proportional to $E_0$ , i.e., to $\sigma$. Thus, ($\sigma – \sigma_p)$ is proportional to $\sigma$ and we can write
$\sigma-\sigma_i=\frac{\sigma}{K}$ (4)
where K is a constant characteristic of the dielectric. Clearly, K > 1. We then have
$V=\frac{\sigma d}{\epsilon_0 K}=\frac{Qd}{A\epsilon_0 A}$ (5)
The capacitance C, with dielectric between the plates, is then
$C=\frac{Q}{V}=\frac{\epsilon_0 KA}{d}$ (6)
The product $\epsilon_0K$ is called the permittivity of the medium and is denoted by $\epsilon$
$\epsilon=\epsilon_0 K$ (7)
For vacuum $K = 1$ and $\epsilon = \epsilon_0$; $\epsilon_0$ is called the permittivity of the vacuum. The dimensionless ratio
$K=\frac{\epsilon}{\epsilon_0}$ (8)
is called the dielectric constant of the substance. As remarked before, from Eq. (4), it is clear that K is greater than 1. From Eqs. (1) and (6)
$K=\frac{C}{C_0}$ (9)
Thus, the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value, when the dielectric is inserted fully between the plates of a capacitor. Though we arrived at Eq. (9) for the case of a parallel plate capacitor, it holds good for any type of capacitor and can, in fact, be viewed in general as a definition of the dielectric constant of a substance.
The dielectric constant is a pure number. Because is always greater than is always greater than unity. Some representative values of are given in Table 1. For vacuum, by definition. For air at ordinary temperatures and pressures, is about 1.0006; this is so nearly equal to 1 that for most purposes an air capacitor is equivalent to one in vacuum. Note that while water has a very large value of it is usually not a very practical dielectric for use in capacitors. The reason is that while pure water is a very poor conductor, it is also an excellent ionic solvent. Any ions that are dissolved in the water will cause charge to flow between the capacitor plates, so the capacitor discharges.
Table 1: Here is the rewritten table of dielectric constant values K at 20∘C:
Thus, we see that a dielectric provides the following advantages:
- Increase in capacitance
- Increase in maximum operating voltage
- Possible mechanical support between the plates, which allows the plates to be close together without touching, thereby decreasing d and increasing C.
Types of Capacitors
Commercial capacitors are often made from metallic foil interlaced with thin sheets of either paraffin-impregnated paper or Mylar as the dielectric material. These alternate layers of metallic foil and dielectric are rolled into a cylinder to form a small package (Fig. 4a). High-voltage capacitors commonly consist of a number of interwoven metallic plates immersed in silicone oil (Fig. 4b). Small capacitors are often constructed from ceramic materials.
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| Fig.4:Three commercial capacitor designs. (a) A tubular capacitor, whose plates are separated by paper and then rolled into a cylinder. (b) A high-voltage capacitor consisting of many parallel plates separated by insulating oil. (c) An electrolytic capacitor. |
Often, an electrolytic capacitor is used to store large amounts of charge at relatively low voltages. This device, shown in Figure 4c, consists of a metallic foil in contact with an electrolyte—a solution that conducts electricity by virtue of the motion of ions contained in the solution. When a voltage is applied between the foil and the electrolyte, a thin layer of metal oxide (an insulator) is formed on the foil, and this layer serves as the dielectric. Very large values of capacitance can be obtained in an electrolytic capacitor because the dielectric layer is very thin, and thus the plate separation is very small.
Electrolytic capacitors are not reversible as are many other capacitors—they have a polarity, which is indicated by positive and negative signs marked on the device. When electrolytic capacitors are used in circuits, the polarity must be aligned properly. If the polarity of the applied voltage is opposite that which is intended, the oxide layer is removed and the capacitor conducts electricity instead of storing charge.
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| Fig.5: A variable capacitor. When one set of metal plates is rotated so as to lie between a fixed set of plates, the capacitance of the device changes. |
Variable capacitors (typically 10 to 500 pF) usually consist of two interwoven sets of metallic plates, one fixed and the other movable, and contain air as the dielectric (Fig. 5). These types of capacitors are often used in radio tuning circuits.
Example 1:
A parallel-plate capacitor has capacitance $C_0=5.00 \ pF$ when there is air between the plates. The separation between the plates is 1.50 mm. (a) What is the maximum magnitude of charge $Q$ that can be placed on each plate if the electric field in the region between the plates is not to exceed $3.00 \times 10^4 \ V/m$? (b) A dielectric with K = 2.70 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed $3.00 \times 10^4 \ V/m$?
Solution:
Given:
Capacitance without the dielectric: $C_0=5.00 \ pF=5.00 \times 10^{-12} \ F$
Separation between the plates: $d= 1.50 \ mm=1.50 \times 10^{-3} \ m$
Maximum electric field: $E_{max}=3.00 \times 10^4 \ V/m$
(a) Maximum charge without the dielectric
We use the following relations:
The relationship between electric field and voltage:
$V_{max}=E_{max}.d$
where $V_{max}$ is the maximum voltage, $E_{max}$ is the maximum electric field, and d is the separation between the plates.
Substituting the values:
$V_{max}=(3.00 \times 10^4 \ V/m)(1.50 \times 10^{-3} \ m)$
$V_{max}=45.0 \ V$
Now, we use the relation between charge, capacitance, and voltage:
$Q=C_0.V_{max}$
Substituting the values:
$Q=(5.00 \ pF=5.00 \times 10^{-12} \ F)(45.0 \ V)$
$Q=2.25 \times 10^{-10} \ C$
So, the maximum charge that can be placed on each plate without exceeding the electric field is: $Q=2.25 \times 10^{-10} \ C$.
(b) Maximum charge with the dielectric
When a dielectric with $K=2.70$ is inserted, the capacitance increases by a factor of K, and the maximum charge will also increase.
The new capacitance with the dielectric is:
$C=K.C_0$
$C=(2.70)(5.00 \ pF=5.00 \times 10^{-12} \ F)=1.35 \times 10^{-11} \ F)$
The electric field in the dielectric is still $E_{max}$, but now the voltage will be different due to the increased capacitance. We can calculate the new voltage with the dielectric:
$V_{max}=E_{max}.d$
(Note that this voltage is the same as in part (a) because the field and the separation are the same.)
Finally, the maximum charge on each plate with the dielectric is:
$V_{max}=45.0 \ V$
Finally, the maximum charge on each plate with the dielectric is:
$Q=C.V_{max}$
Substituting the values:
$Q=(1.35 \times 10^{-11} \ F)(45.0 \ V)$
$Q=6.08 \times 10^{-10} \ C$
So, the maximum charge on each plate with the dielectric inserted is: $Q=6.08 \times 10^{-10} \ C$
Example 2:
Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $E =3.20 \times 10^5 \ V/m$. When the space is filled with dielectric, the electric field is $E =2.50 \times 10^5 \ V/m$. (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?
Solution:
Given:
Electric field in the vacuum: $E_0=3.20 \times 10^5 \ V/m$
Electric field with dielectric: $E=2.50 \times 10^5 \ V/m$
(a) Charge density on each surface of the dielectric
When the space is evacuated, the electric field is related to the charge density on the plates. The electric field between two parallel plates with charge density $\sigma$ is given by the relation:
$E_0=\frac{\sigma}{\epsilon_0}$
From this, we can solve for $\sigma$ (the charge density):
$\sigma=\epsilon_0.E_0$
Substituting the given values:
$\sigma=(8.85 \times 10^{-12} \ F/m)(3.20 \times 10^5 \ V/m)$
$\sigma=2.832 \times 10^{-6} \ C/m^2$
So, the charge density on the plates is $2.832 \times 10^{-6} \ C/m^2$
Now, when the dielectric is inserted, the charge density on the surface of the dielectric is the same as the charge density on the plates. However, the electric field in the dielectric is reduced due to the dielectric constant K. The relation between the electric field in the vacuum $E_0$ and the electric field in the dielectric $E$ is:
$E=\frac{E_0}{K}→K=\frac{E_0}{E}$
Substitute the given values:
$K=\frac{3.20 \times 10^5 \ V/m}{2.50 \times 10^5 \ V/m}=1.28$
Thus, the dielectric constant K is 1.28.
Example 3
Suppose the parallel plates in Fig. 3 each have an area of $2000 \ cm^2$ ($2.00 \times 10^{-1} \ m^2$) and are 1.00 cm ($1.00 \times 10^{-2} \ m$) apart. We connect the capacitor to a power supply, charge it to a potential difference $V_0=3.00 \ kV$ and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1.00 kV while the charge on each capacitor plate remains constant. Find (a) the original capacitance $C_0$, (b) the magnitude of charge $Q$ on each plate; (c) the capacitance $C$ after the dielectric is inserted; (d) the dielectric constant $K$ of the dielectric; (e) the permittivity $\epsilon_0$ of the dielectric; (f ) the magnitude of the induced charge $Q_i$ on each face of the dielectric; (g) the original electric $E_0$ field between the plates; and (h) the electric field $E$ after the dielectric is inserted.
Solution
(a) With vacuum between the plates, we use Eq. $C_0=\frac{\epsilon_0 A}{d}$ with $K=1$:
$C_0=\frac{\epsilon_0 A}{d}=\frac{(8.85 \times 10^{-12} \ F/m)(2.00 \times 10^{-1} \ m^2)}{(1.00 \times 10^{-2} \ m^2)}$
$C_0=1.77 \times 10^{-10} \ F=177 \ pF$
(b) From the definition of capacitance, Eq. (24.1),
$Q=C_0V_0=(1.77 \times 10^{-10} \ F)(3.00 \times 10^2 \ V)$
$Q=5.31 \times 10^{-7} \ C=0.531 \ \mu C$
(c) When the dielectric is inserted, Q is unchanged but the potential difference decreases to V =1.00 kV. Hence from Eq. $C=\frac{Q}{V}$, the new capacitance is
$C=\frac{Q}{V}=\frac{5.31 \times 10^{-7} \ C}{1.00 \times 10^3 \ V}$
$C=5.31 \times 10^{-10} \ F=531 \ pF$
(d) From Eq. $K=\frac{C}{C_0}$, the dielectric constant is
$K=\frac{C}{C_0}=\frac{5.31 \times 10^{-10} \ F}{1.77 \times 10^{-10} \ F}=3.00$
Alternatively, from Eq. $K=\frac{C}{C_0}$,
$K=\frac{V_0}{V}=\frac{3000 \ V}{1000 \ V}=3.00$
(e) Using from part (d) in Eq. $\epsilon=K\epsilon_0$, the permittivity is
$\epsilon=K\epsilon_0=(3.00)(8.85 \times 10^{-12} \ F/m)$
$\epsilon=2.66 \times 10^{-11} \ C^2/N.m^2$
(f) Multiplying both sides of Eq. $Q_i=Q\left(1-\frac{1}{K}\right)$ by the plate area A gives the induced charge $Q_i=\sigma_i A$ in terms of the charge $Q=\sigma A$ on each plate:
$Q_i=Q\left(1-\frac{1}{K}\right)=(5.31 \times 10^{-7} \ C)\left(1-\frac{1}{3.00}\right)$
$Q_i=3.54 \times 10^{-7} \ C$
(g) Since the electric field between the plates is uniform, its magnitude is the potential difference divided by the plate separation:
$E_0=\frac{V_0}{d}=\frac{3000 \ V}{1.00 \times 10^{-2} \ m}=3.00 \times 10^5 \ V/m$
(h) After the dielectric is inserted,
$E=\frac{V}{d}=\frac{1000 \ V}{1.00 \times 10^{-2} \ m}=1.00 \times 10^5 \ V/m$
or, from Eq.,
$E=\frac{\sigma}{\epsilon_0}=\frac{Q}{\epsilon_0 A}$
$E=\frac{(5.31 \times 10^{-7} \ C)}{(2.66 \times 10^{-11} \ C^2/N.m^2)(2.00 \times 10^{-1} \ m^2)}$
$E=1.00 \times 10^5 \ V/m$
or, from Eq.,
$E=\frac{\sigma - \sigma_i}{\epsilon_0}=\frac{Q-Q_i}{\epsilon_0 A}$
$E=\frac{(5.31-3.54) \times 10^{-7} \ C}{(8.85 \times 10^{-12} \ F/m)(2.00 \times 10^{-1} \ m^2)}$
$E=1.00 \times 10^5 \ V/m$
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