Capacitance and Dielectrics Chapter 26 HW: P: 10,18,21,29,33,48, 51,53,54,68
Capacitors Capacitors are devices that store electric charge and energy Examples of where capacitors are used include: radio receivers filters in power supplies energy-storing devices in electronic flashes
Some Uses of Capacitors Defibrillators When fibrillation occurs, the heart produces a rapid, irregular pattern of beats A fast discharge of electrical energy through the heart can return the organ to its normal beat pattern In general, capacitors act as energy reservoirs that can be slowly charged and then discharged quickly to provide large amounts of energy in a short pulse
Capacitance The capacitance, C, of a capacitor is defined as the ratio of the magnitude of the charge on either conductor to the potential difference between the conductors C Q = ΔV The capacitance is a measure of the capacitor s ability to store charge The SI unit of capacitance is the farad (F)
Capacitance The capacitance is proportional to the area of its plates and inversely proportional to the distance between the plates C Q = ΔV C Q Q Q Q εoa = = = = = Δ V Ed σ d ( Q/ ε ) oad d ε o C ε A = o d
Makeup of a Capacitor A capacitor consists of two conductors These conductors are called plates When the conductor is charged, the plates carry charges of equal magnitude and opposite directions A potential difference exists between the plates due to the charge Capacitance will always be a positive quantity The capacitance of a given capacitor is constant The farad is a large unit, typically you will see microfarads (μf) and picofarads (pf) A capacitor stores electrical energy
Parallel Plate Assumptions The assumption that the electric field is uniform is valid in the central region, but not at the ends of the plates If the separation between the plates is small compared with the length of the plates, the effect of the non-uniform field can be ignored The charge density on the plates is σ = Q/A A is the area of each plate, which are equal Q is the charge on each plate, equal with opposite signs
Charging a Parallel Plate Capacitor Each plate is connected to a terminal of the battery If the capacitor is initially uncharged, the battery establishes an electric field in the connecting wires
Active Figure 26.4 AF_2604.html
Energy in a Capacitor Consider the circuit to be a system Before the switch is closed, the energy is stored as chemical energy in the battery When the switch is closed, the energy is transformed from chemical to electric potential energy
Energy Stored in a Capacitor Assume the capacitor is being charged and, at some point, has a charge q on it The work needed to transfer a charge from one plate to the other is The total work required is q dw =Δ Vdq = dq C Q q 0 2 Q 2 C The work done in charging the capacitor appears as electric potential energy U: W = dq = C 2 Q 1 1 ( ) 2 U = = QΔ V = C ΔV 2C 2 2
Energy Stored in a Capacitor 2 Q 1 1 ( ) 2 U = = QΔ V = C ΔV 2C 2 2 This applies to a capacitor of any geometry The energy stored increases as the charge increases and as the potential difference increases In practice, there is a maximum voltage before discharge occurs between the plates The energy can be considered to be stored in the electric field For a parallel-plate capacitor, the energy can be expressed in terms of the field as U = ½ (ε o Ad)E 2 It can also be expressed in terms of the energy density (energy per unit volume) u E = ½ ε o E 2
Dielectrics A dieletric is an insulator that increases the capacitance. C = κε d 0 A Reduced E field prevents breakdown & discharge between plates.
Dielectrics The dieletric constant is the ratio of the field magnitude without and with the dielectric. κ = E 0 E κ κ κ Air Paper Water ~1 = 3.7 = 80 Reduced E field prevents breakdown & discharge between plates.
Dielectrics If the field becomes too great, the dielectric breaks down and becomes a conductor and the plates discharge. Reduced E field prevents breakdown & discharge between plates.
Dielectric Breakdown of Air
Problem What is the maximum voltage that can be sustained between 2 parallel plates separated by 2.5 cm of dry air? Dry air supports max field strength of 3x 10 6 V/m. V = Ed = (3x10 6 V / m)(.025 m) = 7.5x10 4 V = 75kV More than this and the air breaks down and becomes a conductor. LIGHTENING!
The dieletric constant is the ratio of the net electric field magnitude without the dielectric,e 0, and with the dielectric, E. The dielectric reduces the net electric field and potential difference across the plates. E 0 κ = E E = E 0 / κ Reduced E field prevents breakdown & discharge between plates.
Dielectrics An Atomic View The molecules that make up the dielectric are modeled as dipoles The molecules are randomly oriented in the absence of an electric field
Dielectrics An Atomic View An external electric field is applied which produces a torque on the molecules The molecules partially align with the electric field The degree of alignment of the molecules with the field depends inversely upon temperature and directly with the magnitude of the field The degree of alignment of the molecules with the field depends on the polarization of the molecules. E = σ / ε 0 0
Polar vs. Nonpolar Molecules Molecules are said to be polarized when a separation exists between the average position of the negative charges and the average position of the positive charges Polar molecules are those in which this condition is always present Molecules without a permanent polarization are called nonpolar molecules
Water Molecules A water molecule is an example of a polar molecule The center of the negative charge is near the center of the oxygen atom The x is the center of the positive charge distribution The average positions of the positive and negative charges act as point charges Therefore, polar molecules can be modeled as electric dipoles
Induced Polarization A symmetrical molecule has no permanent polarization (a) Polarization can be induced by placing the molecule in an electric field (b) Induced polarization is the effect that predominates in most materials used as dielectrics in capacitors Induced polar molecules can also be modeled as electric dipoles
Electric Dipole An electric dipole consists of two charges of equal magnitude and opposite signs The charges are separated by 2a The electric dipole moment (p) is directed along the line joining the charges from q to +q The electric dipole moment has a magnitude of p = 2aq
Electric Dipole in an External E Field Each charge has a force of F = Eq acting on it The net force on the dipole is zero The forces produce a net torque on the dipole that makes it align with the external electric field. The magnitude of the torque is: τ = 2(Fa sin θ) = 2Eqa sin θ = pe sin θ The torque can also be expressed as the cross product of the moment and the field: τ = p x E
Dielectrics An Atomic View An external field can polarize the dielectric whether the molecules are polar or nonpolar The charged edges of the dielectric act as a second pair of plates producing an induced electric field in the direction opposite the original electric field, thus reducing the net electric field in the dielectric: E = σ / ε 0 0 E = σ / ε ind ind E = E E E = E 0 / κ 0 ind 0
Dielectrics An Atomic View E = E E 0 ind E = σ / ε 0 0 σ σ = σ ind κε ε ε 0 0 0 σ ind = κ 1 κ σ E = σ / ε ind ind 0 κ > 1 σind < σ E = E 0 / κ
Induced Charge and Field E = E E 0 ind The electric field due to the plates is directed to the right and it polarizes the dielectric The net effect on the dielectric is an induced surface charge that results in an induced electric field If the dielectric were replaced with a conductor, the net field between the plates would be zero
Circular Plates C = κε d 0 A C = κc 0 Δ V =ΔV 0 / κ (plate radius = 10 cm) ε 0 = 8.85x10 12 Farad/m
Capacitance of a Cylindrical Capacitor From Gauss s Law, the field between the cylinders is E = 2k e λ / r ΔV = -2k e λ ln (b/a) The capacitance becomes C 0 Q l = = Δ 0 2 ln / 0 V k ( b a ) ΔV 2k ln ( b/ a) e C Q = C = = e l
Coaxial Cable Radio-grade flexible coaxial cable. A: outer plastic sheath B: copper screen C: inner dielectric insulator D: copper core
Geometry of Some Capacitors
Capacitors in Parallel The total charge is equal to the sum of the charges on the capacitors Q total = Q 1 + Q 2 The potential difference across the capacitors is the same And each is equal to the voltage of the battery: ΔV = ΔV 1 = ΔV 2
Capacitors in Parallel The capacitors can be replaced with one capacitor with a capacitance of C eq Q = C ΔV total eq Q total = Q 1 + Q 2 C Δ V = CΔ V + C ΔV eq 1 2 C eq = C 1 + C 2 + Essentially, the areas are combined
Capacitors in Series When a battery is connected to the circuit, electrons are transferred from the left plate of C 1 to the right plate of C 2 through the battery As this negative charge accumulates on the right plate of C 2, an equivalent amount of negative charge is removed from the left plate of C 2, leaving it with an excess positive charge All of the right plates gain charges of Q and all the left plates have charges of +Q thus: Series: Q = Q = Q 1 2 C Q = ΔV
Capacitors in Series The potential differences add up to the battery voltage ΔV = V 1 + V 2 + 1 1 1 = + +K C C C eq 1 2 Δ V = Q C eq The equivalent capacitance of a series combination is always less than any individual capacitor in the combination
Capacitor Summary Capacitors in Parallel: Δ V = Δ V = ΔV Q = Q + Q tot C = C + C +K eq Capacitors in Series: Δ V = Δ V +Δ V + Q 1 2 1 2 1 1 1 = + +K C C C eq 1 2 1 2 1 2 = Q = Q 1 2
Equivalent Capacitance
P22. Three capacitors are connected to a battery as shown in Figure P26.22. Their capacitances are C1 = 3C, C2 = C, and C3 = 5C. (a) What is the equivalent capacitance of this set of capacitors? (b) State the ranking of the capacitors according to the charge they store, from largest to smallest. (c) Rank the capacitors according to the potential differences across them, from largest to smallest. Capacitors in Parallel: Δ V = Δ V = ΔV 1 2 Q = Q + Q tot 1 2 C = C + C +K eq 1 2 Capacitors in Series: Δ V = Δ V +Δ V + Q = Q = Q 1 2 1 2 1 1 1 = + +K C C C eq 1 2 Fig P26-22, p.824
You Try: 27. Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure P26.27. Take C1 = 5.00 μf, C2 = 10.0 μf, and C3 = 2.00 μf. C C C C s p1 p2 eq ( ) ( ) 1 1 1 = + = 3.33 μf 5.00 10.0 = 2 3.33 + 2.00 = 8.66 μf = 2 10.0 = 20.0 μf 1 1 1 = + = 8.66 20.0 6.04 μf Fig P26-27, p.824
P26.72 Assume a potential difference across a and b, and notice that the potential difference across the 8.00 μ F capacitor must be zero by symmetry. Then the equivalent capacitance can be determined from the following circuit: FIG. P26.72
You Try: P26.75 By symmetry, the potential difference across 3C is zero, so the circuit reduces to 1 1 1 8 4 Ceq = + = C = C 2C 4C 6 3.
Variable Capacitors Variable capacitors consist of two interwoven sets of metallic plates One plate is fixed and the other is movable These capacitors generally vary between 10 and 500 pf Used in radio tuning circuits