Capacitance & Dielectrics Chapter 24 opener. Capacitors come in a wide range of sizes and shapes, only a few of which are shown here. A capacitor is basically two conductors that do not touch, and which therefore can store charge of opposite sign on its two conductors. Capacitors are used in a wide variety of circuits, as we shall see in this and later Chapters. Various Capacitors
Capacitors - Definition Capacitor Any configuration of 2 conductors that are close but not touching. A Capacitor has the ability to store electric charge. A “Parallel Plate” Capacitor Figure 24-1. Capacitors: diagrams of (a) parallel plate, (b) cylindrical (rolled up parallel plate).
Makeup of a Capacitor A capacitor always consists of 2 conductors. These are called plates. When the capacitor is charged, the plates carry charges of equal magnitude & opposite sign. A potential difference exists between the plates due to the charge.
Parallel Plate Capacitor Each plate is connected to a battery terminal. The battery is a source of potential difference. If the capacitor is initially uncharged, the battery establishes an electric field in the wires. This field applies a force on electrons in the wire just outside of the plates. The force causes the electrons to move onto the negative plate.
Parallel Plate Capacitor, Continued This process continues until equilibrium is achieved. The plate, the wire & the terminal are then all at the same potential. At this point, there is no field in the wire & the movement of the electrons stops. The plate is now negatively charged. A similar process occurs at the other plate, electrons moving away from the plate & leaving it positively charged. In its final configuration, the potential difference across the capacitor plates is the same as between the battery terminals.
(a) Parallel-Plate Capacitor connected to a battery. (b) A Capacitor in a circuit diagram. Parallel Plate Capacitor Capacitor in a circuit Figure 24-2. (a) Parallel-plate capacitor connected to a battery. (b) Same circuit shown using symbols.
Experiment Shows That 1 F = 1 C/V Capacitance C: When a capacitor is connected to a battery, the charge Q on its plates is proportional to the battery voltage V, with the proportionality constant equal to the Capacitance C: This is The Definition of Capacitance. The SI unit of capacitance is the Farad (F) 1 F = 1 C/V
Definition of Capacitance As we just said, the capacitance, C, of a capacitor is the ratio of the magnitude of the charge on one plate to the potential difference between the plates. As we also said, the SI capacitance unit is The Farad (F) The farad is a large unit, typically you will see microfarads (μF) & picofarads (pF). Capacitance is: Always a positive quantity. Constant for a given capacitor. A measure of the capacitor’s ability to store charge The amount of charge the capacitor can store per unit of potential difference.
Calculating Capacitance: General Procedure or “Recipe” For a given geometry 1. Let the plates (A & B) have charges Q & -Q. 2. Calculate the electric field E between the two plates using techniques of previous chapters 3. Calculate the potential difference V between the plates using (parallel plate capacitor). 4. Use C = Q/V to calculate the capacitance. Figure 24-4. Parallel-plate capacitor, each of whose plates has area A. Fringing of the field is ignored.
Example: Parallel Plate Capacitor Earlier Result: The magnitude of the electric field E between 2 closely space charged plates with charge Q & area A is For plates of area A, the surface charge density is: σ = (Q/A). So, E between the plates is: E = Q/(ε0A). The relation between V & E: Combining these gives the potential difference VB - VA = (Qd)/(ε0A). Figure 24-4. Parallel-plate capacitor, each of whose plates has area A. Fringing of the field is ignored. So, the capacitance of a parallel plate capacitor is:
Example: Parallel Plate Capacitor The charge density on the plates σ = (Q/A). A is the area of each plate. Q is the charge on each plate, equal with opposite signs. The electric field is uniform between the plates & zero elsewhere. So, the capacitance is proportional to the area of its plates & inversely proportional to the distance between the plates.
Example: Calculate (a) The capacitance C of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm & are separated by a 1.0-mm air gap. Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
C = 53 pF Example: Calculate Answer (a) The capacitance C of a parallel-plate capacitor whose plates are 20 cm × 3.0 cm & are separated by a 1.0-mm air gap. Answer Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2. C = 53 pF
Example: Calculate (b) The charge Q on each plate if a 12-V battery is connected across the 2 plates. Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
Answer Q = CV = 6.4 10-10 C Example: Calculate (b) The charge Q on each plate if a 12-V battery is connected across the 2 plates. Answer Q = CV = 6.4 10-10 C Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
Example: Calculate (c) The electric field E between the plates. Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
Answer E = V/d = 1.2 104 V/m Example: Calculate (c) The electric field E between the plates. Answer E = V/d = 1.2 104 V/m Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
Example: Calculate (d) An estimate of the area A of the plates needed to achieve a capacitance of C = 1 F, given the same air gap d. Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
A = (Cd/ε0) = 108 m2 !!! = 100 km2 36 square miles!! Example: Calculate (d) An estimate of the area A of the plates needed to achieve a capacitance of C = 1 F, given the same air gap d. Answer A = (Cd/ε0) = 108 m2 !!! = 100 km2 36 square miles!! Solution: a. C = 53 pF. b. Q = CV = 6.4 x 10-10 C. c. E = V/d = 1.2 x 104 V/m. d. A = Cd/ε0 = 108 m2.
Capacitors can now be made with capacitances of C = 1 F or more, but these are NOT parallel-plate capacitors. They are usually made from activated carbon, which acts a capacitor on a small scale. The capacitance of 0.1 g of activated carbon is about 1 F. Some computer keyboards use capacitors; depressing the key changes the capacitance, which is detected in a circuit Figure 24-5. Key on a computer keyboard. Pressing the key reduces the capacitor spacing thus increasing the capacitance which can be detected electronically.
Example: Cylindrical Capacitor See figure. A Cylindrical Capacitor consists of a cylinder (or wire) of radius Rb surrounded by a coaxial cylindrical shell of inner radius Ra. Both cylinders have length ℓ, which is assumed to be much greater than the separation of the cylinders, so “end effects” can be neglected. The capacitor is charged (by connecting it to a battery) so that one cylinder has charge +Q (say, the inner one) the other one charge -Q. Formula for the capacitance C (using calculus) is: Figure 24-6. (a) Cylindrical capacitor consists of two coaxial cylindrical conductors. (b) The electric field lines are shown in cross-sectional view. Solution: We need to find the potential difference between the cylinders; we can do this by integrating the field (which was calculated for a long wire already). The field is proportional to 1/R, so the potential is proportional to ln Ra/Rb. Then C = Q/V.
Capacitance of Cylindrical Capacitor Results Notation Change from previous slide! Rb a, Ra b Sorry! The potential difference is DV = -2keln (b/a) = (Q/ℓ) So the capacitance is
Example: Spherical Capacitor See Figure A Spherical Capacitor consists of 2 concentric spherical, conducting shells of radii ra & rb. The inner shell carries a uniformly distributed charge Q on its surface. The outer shell carries an equal & opposite charge -Q. Formula for the capacitance C (using calculus) is: Figure 24-7. Cross section through the center of a spherical capacitor. The thin inner shell has radius rb and the thin outer shell has radius ra. Solution: As in the previous example, we find the potential difference by integrating the field (that of a point charge) from rb to ra. Then C = Q/V.
Capacitance of a Spherical Capacitor Results Notation Change from previous slide! rb a, ra b Sorry! The potential difference is: l So, the capacitance is:
Capacitance of 2 Long Parallel Wires Example: Capacitance of 2 Long Parallel Wires Estimate the capacitance per unit length C/ℓ of 2 very long straight parallel wires, each of radius R, carrying uniform charges +Q & –Q, separated by a distance d which is large compared to R (d >> R). The line charge density is λ = Q/ℓ Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.
Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law: At x from wire a: Ea = [λ/(20x). Note again: λ = Q/ℓ At (d – x) from wire b: Eb = [λ/(20{d – x})] Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.
Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law: At x from wire a: Ea = [λ/(20x). Note again: λ = Q/ℓ At (d – x) from wire b: Eb = [λ/(20{d – x})] Total: E = Ea + Eb Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.
Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law: At x from wire a: Ea = [λ/(20x). Note again: λ = Q/ℓ At (d – x) from wire b: Eb = [λ/(20{d – x})] Total: E = Ea + Eb Potential Difference: V = Vb – Va = - (Eb - Ea)x. Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.
Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law: At x from wire a: Ea = [λ/(20x). Note again: λ = Q/ℓ At (d – x) from wire b: Eb = [λ/(20{d – x})] Total: E = Ea + Eb Potential Difference: V = Vb – Va = - (Eb - Ea)x.. Limit: R d – R. Simplifying with the assumption: d >> R gives: V [Q/(0ℓ )] ln(d/R). Capacitance: C = Q/V so C (0ℓ)/ln(d/R) Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.
Capacitance of 2 Long Parallel Wires E fields from Gauss’s Law: At x from wire a: Ea = [λ/(20x). Note again: λ = Q/ℓ At (d – x) from wire b: Eb = [λ/(20{d – x})] Total: E = Ea + Eb Potential Difference: V = Vb – Va = - (Eb - Ea)x. Limit: R d – R. Simplifying with the assumption: d >> R gives: V [Q/(0ℓ )] ln(d/R). Capacitance: C = Q/V so C (0ℓ)/ln(d/R) Solution: The field between the wires is the superposition of the fields due to each wire. Integrating gives the potential; then C = Q/V. See text for details. Since the charge is a charge density (charge per unit length), the capacitance is also quoted as C/L – capacitance per unit length.