Chapter 24 Capacitance, Dielectrics, Electric Energy Storage 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.

24-4 Electric Energy Storage A charged capacitor stores electric energy; the energy stored is equal to the work done to charge the capacitor:

24-4 Electric Energy Storage Conceptual Example 24-9: Capacitor plate separation increased. A parallel-plate capacitor carries charge Q and is then disconnected from a battery. The two plates are initially separated by a distance d. Suppose the plates are pulled apart until the separation is 2d. How has the energy stored in this capacitor changed? Solution: Increasing the plate separation decreases the capacitance, but the charge remains the same. Therefore the energy, U = ½ Q2/C, doubles.

24-4 Electric Energy Storage The energy density, defined as the energy per unit volume, is the same no matter the origin of the electric field: The sudden discharge of electric energy can be harmful or fatal. Capacitors can retain their charge indefinitely even when disconnected from a voltage source – be careful!

24-4 Electric Energy Storage Heart defibrillators use electric discharge to “jump-start” the heart, and can save lives. Figure 24-14. Heart defibrillator.

24-4 Electric Energy Storage National Ignition Facility (NIF) Lawrence Livermore National Laboratory Solution: a. U = ½ CV2 = 3.0 J. b. P = U/t = 3000 W.

NIF Laser system driven by 4000 300 μF capacitors which store a total of 422 MJ. They take 60 s to charge and are discharged in 400 μs. What is the potential difference across each capacitor? What is the power delivered during the discharge?

NIF Laser system driven by 4000 300 μF capacitors which store a total of 422 MJ. They take 60 s to charge and are discharged in 400 μs. What is the potential difference across each capacitor? What is the power delivered during the discharge? Solution: U = CV2/2 →V = (2U/C)1/2 →V = [2(422x106)/4000/300x10-6] ½ = 26.5 kV P = W/t = U/t = 422x106 /400x10-6 ~ 1012 W = 1000 GW! cf. 1.0-1.5 GW for power plant

24-6 Molecular Description of Dielectrics The molecules in a dielectric, when in an external electric field, tend to become oriented in a way that opposes the external field. A dielectric is an insulator, and is characterized by a dielectric constant K. Figure 24-17. Molecular view of the effects of a dielectric.

24-6 Molecular Description of Dielectrics This means that the electric field within the dielectric is less than it would be in air, allowing more charge to be stored for the same potential. This reorientation of the molecules results in an induced charge – there is no net charge on the dielectric, but the charge is asymmetrically distributed. The magnitude of the induced charge depends on the dielectric constant:

24-5 Dielectrics Capacitance of a parallel-plate capacitor filled with dielectric: Using the dielectric constant, we define the permittivity:

24-5 Dielectrics Dielectric strength is the maximum field a dielectric can experience without breaking down.

24-5 Dielectrics Here are two experiments where we insert and remove a dielectric from a capacitor. In the first, the capacitor is connected to a battery, so the voltage remains constant. The capacitance increases, and therefore the charge on the plates increases as well. Figure 24-15a. Dielectric inserted into a capacitor, voltage held constant.

24-5 Dielectrics In this second experiment, we charge a capacitor, disconnect it, and then insert the dielectric. In this case, the charge remains constant. Since the dielectric increases the capacitance, the potential across the capacitor drops. Figure 24-15b. Dielectric inserted into a capacitor, charge held constant.

24-5 Dielectrics Example 24-11: Dielectric removal. A parallel-plate capacitor, filled with a dielectric with K = 3.4, is connected to a 100-V battery. After the capacitor is fully charged, the battery is disconnected. The plates have area A = 4.0 m2 and are separated by d = 4.0 mm. (a) Find the capacitance, the charge on the capacitor, the electric field strength, and the energy stored in the capacitor. (b) The dielectric is carefully removed, without changing the plate separation nor does any charge leave the capacitor. Find the new values of capacitance, electric field strength, voltage between the plates, and the energy stored in the capacitor. Solution: a. C = Kε0A/d = 3.0 x 10-8 F. Q = CV = 3.0 x 10-6 C. E = V/d = 25 kV/m. U = ½ CV2 = 1.5 x 10-4 J. B. Now C = 8.8 x 10-9 F, Q = 3.0 x 10-6 C (no change), V = 340 V, E = 85 kV/m, U = 5.1 x 10-4 J. The increase in energy comes from the work it takes to remove the dielectric.

Summary of Chapter 24 Capacitor: nontouching conductors carrying equal and opposite charge. Capacitance: Capacitance of a parallel-plate capacitor:

Summary of Chapter 24 Capacitors in parallel: Capacitors in series:

Summary of Chapter 24 Energy density in electric field: A dielectric is an insulator. Dielectric constant gives ratio of total field to external field. For a parallel-plate capacitor:

Chapter 25 Electric Currents and Resistance Chapter 25 opener. The glow of the thin wire filament of a light bulb is caused by the electric current passing through it. Electric energy is transformed to thermal energy (via collisions between moving electrons and atoms of the wire), which causes the wire’s temperature to become so high that it glows. Electric current and electric power in electric circuits are of basic importance in everyday life. We examine both dc and ac in this Chapter, and include the microscopic analysis of electric current.

25-1 The Electric Battery Volta discovered that electricity could be created if dissimilar metals were connected by a conductive solution called an electrolyte. This is a simple electric cell. Figure 25-3. Simple electric cell.

25-1 The Electric Battery Several cells connected together make a battery, although now we refer to a single cell as a battery as well. Figure 25-4. (a) Diagram of an ordinary dry cell (like a D-cell or AA). The cylindrical zinc cup is covered on the sides; its flat bottom is the negative terminal. (b) Two dry cells (AA type) connected in series. Note that the positive terminal of one cell pushes against the negative terminal of the other.

Unit of electric current: the ampere, A: Electric current is the rate of flow of charge through a conductor: The instantaneous current is given by: Unit of electric current: the ampere, A: 1 A = 1 C/s.

25-2 Electric Current A complete circuit is one where current can flow all the way around. Note that the schematic drawing doesn’t look much like the physical circuit! Figure 25-6. (a) A simple electric circuit. (b) Schematic drawing of the same circuit, consisting of a battery, connecting wires (thick gray lines), and a lightbulb or other device.

25-2 Electric Current Example 25-1: Current is flow of charge. A steady current of 2.5 A exists in a wire for 4.0 min. (a) How much total charge passed by a given point in the circuit during those 4.0 min? (b) How many electrons would this be? Solution: a. Charge = current x time; convert minutes to seconds. Q = 600 C. b. Divide by electron charge: n = 3.8 x 1021 electrons.

ConcepTest 25.1 Connect the Battery Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 2) 3)

ConcepTest 25.1 Connect the Battery Which is the correct way to light the lightbulb with the battery? 4) all are correct 5) none are correct 1) 2) 3) Current can flow only if there is a continuous connection from the negative terminal through the bulb to the positive terminal. This is the case for only Fig. (3).

25-2 Electric Current By convention, current is defined as flowing from + to -. Electrons actually flow in the opposite direction, but not all currents consist of electrons. Figure 25-8. Conventional current from + to - is equivalent to a negative electron flow from – to +.

25-3 Ohm’s Law: Resistance and Resistors Experimentally, it is found that the current in a wire is proportional to the potential difference between its ends:

25-3 Ohm’s Law: Resistance and Resistors The ratio of voltage to current is called the resistance:

25-3 Ohm’s Law: Resistance and Resistors In many conductors, the resistance is independent of the voltage; this relationship is called Ohm’s law. Materials that do not follow Ohm’s law are called nonohmic. Figure 25-9. Graphs of current vs. voltage for (a) a metal conductor which obeys Ohm’s law, and (b) for a nonohmic device, in this case a semiconductor diode. Unit of resistance: the ohm, Ω: 1 Ω = 1 V/A.

25-3 Ohm’s Law: Resistance and Resistors Conceptual Example 25-3: Current and potential. Current I enters a resistor R as shown. (a) Is the potential higher at point A or at point B? (b) Is the current greater at point A or at point B? Solution: a. Point A is at higher potential (current flows “downhill”). b. The current is the same – all the charge that flows past A also flows past B.

25-3 Ohm’s Law: Resistance and Resistors Example 25-4: Flashlight bulb resistance. A small flashlight bulb draws 300 mA from its 1.5-V battery. (a) What is the resistance of the bulb? (b) If the battery becomes weak and the voltage drops to 1.2 V, how would the current change? Figure 25-11. Flashlight (Example 25–4). Note how the circuit is completed along the side strip. a. R = V/I = 5.0 Ω. b. Assuming the resistance stays the same, the current will drop to 240 mA.

25-3 Ohm’s Law: Resistance and Resistors Some clarifications: Batteries maintain a (nearly) constant potential difference; the current varies. Resistance is a property of a material or device. Current is not a vector but it does have a direction. Current and charge do not get used up. Whatever charge goes in one end of a circuit comes out the other end.

ConcepTest 25.2 Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? 1) Ohm’s law is obeyed since the current still increases when V increases 2) Ohm’s law is not obeyed 3) this has nothing to do with Ohm’s law

Follow-up: Where could this situation occur? ConcepTest 25.2 Ohm’s Law You double the voltage across a certain conductor and you observe the current increases three times. What can you conclude? 1) Ohm’s law is obeyed since the current still increases when V increases 2) Ohm’s law is not obeyed 3) this has nothing to do with Ohm’s law Ohm’s law, V = IR, states that the relationship between voltage and current is linear. Thus, for a conductor that obeys Ohm’s law, the current must double when you double the voltage. Follow-up: Where could this situation occur?

25-4 Resistivity The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area: The constant ρ, the resistivity, is characteristic of the material.

25-4 Resistivity This table gives the resistivity and temperature coefficients of typical conductors, semiconductors, and insulators.

25-4 Resistivity Example 25-5: Speaker wires. Suppose you want to connect your stereo to remote speakers. (a) If each wire must be 20 m long, what diameter copper wire should you use to keep the resistance less than 0.10 Ω per wire? (b) If the current to each speaker is 4.0 A, what is the potential difference, or voltage drop, across each wire? Solution: a. R = ρl/A; you can solve this for A and then find the diameter. D = 2.1 mm. b. V = IR = 0.40 V.

25-4 Resistivity For any given material, the resistivity increases with temperature: Semiconductors are complex materials, and may have resistivities that decrease with temperature.

25-4 Resistivity Example 25-7: Resistance thermometer. The variation in electrical resistance with temperature can be used to make precise temperature measurements. Platinum is commonly used since it is relatively free from corrosive effects and has a high melting point. Suppose at 20.0°C the resistance of a platinum resistance thermometer is 164.2 Ω. When placed in a particular solution, the resistance is 187.4 Ω. What is the temperature of this solution? Solution: The resistance is proportional to the resistivity; use the temperature dependence of resistivity to find the temperature. T = 56.0 °C.

ConcepTest 25.3a Wires I 1) dA = 4dB Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare?

ConcepTest 25.3a Wires I 1) dA = 4dB Two wires, A and B, are made of the same metal and have equal length, but the resistance of wire A is four times the resistance of wire B. How do their diameters compare? The resistance of wire A is greater because its area is less than wire B. Since area is related to radius (or diameter) squared, the diameter of A must be two times less than the diameter of B.

25-5 Electric Power Power, as in kinematics, is the energy transformed by a device per unit time: or

25-5 Electric Power The unit of power is the watt, W. For ohmic devices, we can make the substitutions:

25-5 Electric Power Example 25-8: Headlights. Calculate the resistance of a 40-W automobile headlight designed for 12 V. Solution: R = V2/P = 3.6 Ω.

25-5 Electric Power What you pay for on your electric bill is not power, but energy – the power consumption multiplied by the time. We have been measuring energy in joules, but the electric company measures it in kilowatt-hours, kWh: 1 kWh = (1000 W)(3600 s) = 3.60 x 106 J.

25-5 Electric Power Example 25-9: Electric heater. An electric heater draws a steady 15.0 A on a 120-V line. How much power does it require and how much does it cost per month (30 days) if it operates 3.0 h per day and the electric company charges 9.2 cents per kWh? Solution: P = IV = 1800 W. 1800 W x 3.0 h/day x 30 days = 162 kWh. At 9.2 cents per kWh, this would cost $15.

25-6 Power in Household Circuits Conceptual Example 25-12: A dangerous extension cord. Your 1800-W portable electric heater is too far from your desk to warm your feet. Its cord is too short, so you plug it into an extension cord rated at 11 A. Why is this dangerous? Solution: An 1800-W heater operating at 120 V draws 15 A of current, which exceeds the rating of the extension cord. This creates a risk of overheating and fire.

25-7 Alternating Current Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC). Figure 25-21. (a) Direct current. (b) Alternating current.

25-7 Alternating Current The voltage varies sinusoidally with time: , , as does the current:

25-7 Alternating Current Multiplying the current and the voltage gives the power: Figure 25-22. Power transformed in a resistor in an ac circuit.

25-7 Alternating Current Usually we are interested in the average power: .

25-7 Alternating Current The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root-mean-square (rms) value:

25-7 Alternating Current Example 25-13: Hair dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain? Figure 25-23. A hair dryer. Most of the current goes through the heating coils, a pure resistance; a small part goes to the motor to turn the fan. Example 25–13. Solution: a. The rms current is the power divided by the rms voltage, or 8.33 A, so the peak current is 11.8 A. Then R = V/I = 14.4 Ω. b. Assuming the resistance would stay the same (probably wrong – the heat would increase it), doubling the voltage would mean doubling the current, and the power (P = IV) would increase by a factor of 4. This would not be good for the hair dryer.

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity Electrons in a conductor have large, random speeds just due to their temperature. When a potential difference is applied, the electrons also acquire an average drift velocity, which is generally considerably smaller than the thermal velocity. Figure 25-24. Electric field E in a uniform wire of cross-sectional area A carrying a current I. The current density j = I/A.

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity We define the current density (current per unit area) – this is a convenient concept for relating the microscopic motions of electrons to the macroscopic current: If the current is not uniform: .

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity This drift speed is related to the current in the wire, and also to the number of electrons per unit volume: and

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity Example 25-14: Electron speeds in a wire. A copper wire 3.2 mm in diameter carries a 5.0-A current. Determine (a) the current density in the wire, and (b) the drift velocity of the free electrons. (c) Estimate the rms speed of electrons assuming they behave like an ideal gas at 20°C. Assume that one electron per Cu atom is free to move (the others remain bound to the atom). Solution: a. j = I/A = 6.2 x 105 A/m2 b. The number of free electrons per unit volume is equal to the number of atoms per unit volume. For pure copper, the number of atoms per unit volume can be written as the number of atoms in a mole divided by the volume of a mole of copper; the volume of a mole of copper is given by the atomic mass of copper in grams (63.5 g = 1 mole of copper) divided by the density of copper. This gives 8.4 x 1028 atoms/m3. Then the drift velocity is j/ne = 4.6 x 10-5 m/s. c. If we model the electrons as a three-dimensional single-atom gas, the rms speed is given by the square root of 3kT/m, which is 1.2 x 105 m/s. (This actually underestimates the speed of the electrons by about a factor of 10).

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity The electric field inside a current-carrying wire can be found from the relationship between the current, voltage, and resistance. Writing R = ρ l/A, I = jA, and V = El , and substituting in Ohm’s law gives:

25-8 Microscopic View of Electric Current: Current Density and Drift Velocity Example 25-15: Electric field inside a wire. What is the electric field inside the wire of Example 25–14? (The current density was found to be 6.2 x 105 A/m2.) Solution: E = ρj = 0.01 V/m.

25-9 Superconductivity In general, resistivity decreases as temperature decreases. Some materials, however, have resistivity that falls abruptly to zero at a very low temperature, called the critical temperature, TC. Figure 25-27. A superconducting material has zero resistivity when its temperature is below TC, its “critical temperature.” At TC, the resistivity jumps to a “normal” nonzero value and increases with temperature as most materials do (Eq. 25–5).

25-9 Superconductivity Experiments have shown that currents, once started, can flow through these materials for years without decreasing even without a potential difference. Critical temperatures are low; for many years no material was found to be superconducting above 23 K. Since 1987, new materials have been found that are superconducting below 90 K, and work on higher temperature superconductors is continuing.

Summary of Chapter 25 A battery is a source of constant potential difference. Electric current is the rate of flow of electric charge. Conventional current is in the direction that positive charge would flow. Resistance is the ratio of voltage to current:

Summary of Chapter 25 Ohmic materials have constant resistance, independent of voltage. Resistance is determined by shape and material: ρ is the resistivity.

Summary of Chapter 25 Power in an electric circuit: Direct current is constant. Alternating current varies sinusoidally:

Summary of Chapter 25 The average (rms) current and voltage: Relation between drift speed and current:

Homework Assignment # 5 Chapter 24 – 60, 82 Chapter 25 – 10, 20, 40, 54, 58 Tentative HW # 6: Chapter 26 –