A capacitor is an arrangement of conductors that is used to store electric charge. A very simple capacitor is an isolated metallic sphere. The potential of a sphere with radius R and charge Q is equal to
27.2. The parallel-plate capacitorAnother example of a capacitor is a system consisting of two parallel metallic plates. In Chapter 26 it was shown that the potential difference between two plates of area A, separation distance d, and with charges +Q and -Q, is given by
Example: Problem 27.7The tube of a Geiger counter consists of a thin straight wire surrounded by a coaxial conducting shell. The diameter of the wire is 0.0025 cm and that of the shell is 2.5 cm. The length of the tube is 10 cm. What is the capacitance of a Geiger-counter tube ?
27.3. Capacitors in CombinationThe symbol of a capacitor is shown in Figure 27.2. Capacitors can be connected together; they can be connected in series or in parallel. Figure 27.3 shows two capacitors, with capacitance C1 and C2, connected in parallel. The potential difference across both capacitors must be equal and therefore
Example: Problem 27.10A multi-plate capacitor, such as used in radios, consists of four parallel plates arranged one above the other as shown in Figure 27.5. The area of each plate is A, and the distance between adjacent plates is d. What is the capacitance of this arrangement ?
Example: Problem 27.13Three capacitors, of capacitance C1 = 2.0 uF, C2 = 5.0 uF, and C3 = 7.0 uF, are initially charged to 36 V by connecting each, for a few instants, to a 36-V battery. The battery is then removed and the charged capacitors are connected in a closed series circuit, with the positive and negative terminals joined as shown in Figure 27.7. What will be the final charge on each capacitor ? What will be the voltage across the points PP' ?
27.4. DielectricsIf the space between the plates of a capacitor is filled with an insulator, the capacitance of the capacitor will chance compared to the situation in which there is vacuum between the plates. The change in the capacitance is caused by a change in the electric field between the plates. The electric field between the capacitor plates will induce dipole moments in the material between the plates. These induced dipole moments will reduce the electric field in the region between the plates. A material in which the induced dipole moment is linearly proportional to the applied electric field is called a linear dielectric. In this type of materials the total electric field between the capacitor plates E is related to the electric field Efree that would exist if no dielectric was present:
The potential difference across a capacitor is proportional to the electric field between the plates. Since the presence of a dielectric reduces the strength of the electric field, it will also reduce the potential difference between the capacitor plates (if the total charge on the plates is kept constant):
The electric field between the two capacitor plates is the vector sum of the fields generated by the charges on the capacitor and the field generated by the surface charges on the surface of the dielectric. The electric field generated by the charges on the capacitor plates (charge density of [sigma]free) is given by
Example: Problem 27.19A parallel plate capacitor of plate area A and separation distance d contains a slab of dielectric of thickness d/2 (see Figure 27.8) and dielectric constant [kappa]. The potential difference between the plates is [Delta]V.
a) In terms of the given quantities, find the electric field in the empty region of space between the plates.
b) Find the electric field inside the dielectric.
c) Find the density of bound charges on the surface of the dielectric.
27.5. Gauss Law in DielectricsThe electric field in an "empty" capacitor can be obtained using Gauss' law. Consider an ideal capacitor (with no fringing fields) and the integration volume shown in Figure 27.9. The area of each capacitor plate is A and the charges on the plates are +/-Q. The charge enclosed by the integration volume shown in Figure 27.9 is equal to +Q. Gauss' law states that the electric flux [Phi] through the surface of the integration volume is related to the enclosed charge:
Example: Problem 27.25A metallic sphere of radius R is surrounded by a concentric dielectric shell of inner radius R, and outer radius 3R/2. This is surrounded by a concentric, thin, metallic shell of radius 2R (see Figure 27.10). The dielectric constant of the shell is [kappa]. What is the capacitance of this contraption ?
Suppose the charge on the inner sphere is Qfree. The electric field inside the dielectric can be determined by applying Gauss' law for a dielectric (eq.(27.50)) and using as the integration volume a sphere of radius r (where R < r < 3R/2)
27.6 Energy in CapacitorsThe electric potential energy of a capacitor containing no dielectric and with charge +/-Q on its plates is given by
Example: Problem 27.40Ten identical 5 uF capacitors are connected in parallel to a 240-V battery. The charged capacitors are then disconnected from the battery and reconnected in series, the positive terminal of each capacitor being connected to the negative terminal of the next. What is the potential difference between the negative terminal of the first capacitor and the positive terminal of the last capacitor ? If these terminals are connected via an external circuit, how much charge will flow around this circuit as the series arrangement discharges ? How much energy is released in the discharge ? Compare this charge and this energy with the charge and energy stored in the original, parallel arrangement, and explain any discrepancies.
The charge on each capacitor, after being connected to the 240-V battery, is equal to