Chapter 24 Capacitance and Dielectrics Lecture by Dr. Hebin Li
Goals for Chapter 24 To understand capacitors and calculate capacitance To analyze networks of capacitors To calculate the energy stored in a capacitor To examine dielectrics and how they affect capacitance
Introduction: storing energy in a capacitor + + + + + + -+ + + + + It requires work to move charges between two conductors separated by an insulator or a vacuum. The energy is stored in the system of the two conductors. - - - - - - - - - - A capacitor can be formed by any two conductors separated by an insulator (or a vacuum). Capacitors have important applications in electronics.
Capacitors and capacitance Any two conductors separated by an insulator (or a vacuum) form a capacitor. There is a fixed potential difference for the given charge Q and Q on the conductors. For a capacitor, the ratio of the charge to potential difference is constant. This ratio is called the capacitance C of the capacitor. C = Q V ab The SI unite of capacitance is called farad (1 F). 1 F = 1 C/V One farad is a very large capacitance. We usually use 1 µf, 1 nf, or 1 pf.
Calculating capacitance: parallel-plate capacitor The simplest capacitor consists of two parallel conducting plates. (Parallel-plate capacitor). If the plates have charges Q and Q, the electric field E = σ = Q ε 0 ε 0 A The potential difference between the plates is V ab = Ed = Qd ε 0 A So the capacitance is C = Q A = ε V 0 ab d C = ε 0 A d
A spherical capacitor From Example 23.8, we know that the potential at any point between the spheres is Q V = 4πε 0 r So the potential difference between the inner and outer conductors is The capacitance is
A cylindrical capacitor From Example 23.10, we know that the potential outside a charged cylinder is V = λ ln r 0 2πε 0 r So the potential difference between the inner and outer conductors is The capacitance of a length L is The capacitance per unit length is
Example The plates of a parallel-plate capacitor in vacuum are 5.00 mm apart and 2.00 m2 in area. (a)what is the capacitance? (b)how big the plates would have to be to make it a 1.0-F capacitor? (a) (b) A = Cd ε 0 = 1 F 5.00 10 3 m 8.85 10 12 F/m = 5.6 108 m 2 If the plates are squares, each side would be 23.8 km (14.8 mi)
Capacitors in series The total charge on the lower plate of C 1 and the upper plate of C 2 is zero. The magnitude of charge on all plates is the same. The potential differences on each capacitor: The equivalent capacitor has the potential difference The equivalent capacitance For capacitors in series:
Capacitors in parallel Capacitors in parallel have the same potential difference V. The charge on each capacitor: The charge on the equivalent capacitor is the sum of charges on all capacitors in parallel The equivalent capacitance For capacitors in parallel:
Calculating capacitance: a capacitor network A capacitor network with capacitors that are neither all in series nor all in parallel. The strategy is to combine the capacitors that are in series or in parallel, find their equivalent capacitance, and replace them with the equivalent capacitor. With the equivalent capacitor in place, repeat the process until the network is reduced to a single equivalent capacitor.
Energy stored in a capacitor Assume a capacitor has charge q and potential difference v, then v = q C At this stage, the work dw required to transfer an additional element of charge dq is Now if we consider the process to charge the capacitor from zero to a final charge Q, the total work W needed is This energy is stored in the capacitor. The potential energy stored in a capacitor is Electric energy density:
Example: For the system of capacitors shown in the figure below, a potential difference of 25 V is maintained across ab. (a) What is the equivalent capacitance of this system between a and b? (b) How much charge is stored by this system? (c) How much charge does the 6.5-nF capacitor store? (d) What is the potential difference acros the 7.5-nF capacitor?
Example: For the capacitor network shown in the figure below, the potential difference across ab is 36 V. Find (a) the total charge stored in this network; (b) the charge on each capacitor; (c) the total energy stored in the network; (d) the energy stored in each capacitor; (e) the potential difference across each capacitor.
Example: In the figure below, each capacitance C 1 is 6.9 μf, and each capacitance C 2 is 4.6 μf. (a) Calculate the equivalent capacitance of the network between points a and b. (b) Find the charge on each of the three capacitors nearest a and b when V ab = 420 V. (c) What is V cd when V ab = 420 V?
Dielectrics A dielectric is a nonconducting material. Most capacitors have dielectric between their plates. (See Figure at upper right.) The dielectric constant of the material is K = C/C 0 > 1. Dielectric increases the capacitance and the energy density by a factor K. Figure (lower right) shows how the dielectric affects the electric field between the plates. Table on the next slide shows some values of the dielectric constant.
Table 24.1 Some dielectric constants
Molecular model of induced charge - I Figures 24.17 (right) and 24.18 (below) show the effect of an applied electric field on polar and nonpolar molecules.
Molecular model of induced charge - II Figure below shows polarization of the dielectric and how the induced charges reduce the magnitude of the resultant electric field.
Induced charge and polarization When a dielectric material is inserted between the plates, an induced charge of the opposite sign appears on each surface of the dielectric. The induced charge is a result of redistribution of positive and negative charge within the dielectric materials, a phenomenon called polarization. Without the dielectric: E 0 = σ ε 0 With the dielectric: E = σ σ i ε 0 The induced charge density: σ i = σ 1 1 K The permittivity is defined as ε = Kε 0, then E = σ ε The capacitance with the dielectric is
Example: A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40. An 18.0-V battery is connected across the plates. (a) What is the capacitance of this combination? (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas but change nothing else, how much energy will be stored in the capacitor?