General Physics (PHY 2140) Lecture 4 Electrostatics and electrodynamics Capacitance and capacitors capacitors with dielectrics Electric current current and drift speed resistance and Ohm s law resistivity Last lecture: 1. Equipotential surfaces. Lightning Review They are defined as a surface in space on which the potential is the same for every point (surfaces of constant voltage) The electric field at every point of an equipotential surface is perpendicular to the surface http://www.physics.wayne.edu/~alan/2140website/main.htm Chapter 16-17 1 2 Last lecture: 1. Capacitance and capacitors Combinations of capacitors Parallel Series Parallel-plate capacitor Energy stored in a capacitor Lightning Review Q C = Ceq = C1 C2... 1 1 1 =... Ceq C1 C2 2 1 Q 1 U = QV = = CV C = ε0 d 2 2C 2 Review Problem: Consider an isolated simple parallel-plate capacitor whose plates are given equal and opposite charges and are separated by a distance d. Suppose the plates are pulled apart until they are separated by a distance D>d. The electrostatic energy stored in a capacitor is a. greater than b. the same as c. smaller than before the plates were pulled apart. 3 2 16.10 Capacitors with dielectrics dielectric is an insulating material (rubber, glass, etc.) Consider an insolated, charged capacitor Q Q Q V 0 V Notice that the potential difference decreases (κ = V 0 /V) Since charge stayed the same (Q=Q 0 ) capacitance increases C Q Q κq 0 0 0 = = = = V V0 κ V0 dielectric constant: κ = C/C 0 Dielectric constant is a material property Q κc 0 nsert a dielectric 4 1
Capacitors with dielectrics - notes Capacitance is multiplied by a factor k when the dielectric fills the region between the plates completely E.g., for a parallel-plate plate capacitor C = κε 0 d The capacitance is limited from above by the electric discharge that can occur through the dielectric material separating the plates n other words, there exists a maximum of the electric field, sometimes called dielectric strength,, that can be produced in the dielectric before it breaks down Dielectric constants and dielectric strengths of various materials at room temperature Material Vacuum ir Water Fused quartz Dielectric constant, κ 1.00 1.00059 80 3.78 Dielectric strength (V/m) -- 3 10 6 1.5 10 7 * 9 10 6 For a more complete list, see Table 16.1 * For short duration (μsec) pulses 5 6 Example Take a parallel plate capacitor whose plates have an area of 2.0 m 2 and are separated by a distance of 1mm. The capacitor is charged to an initial voltage of 3 kv and then disconnected from the charging source. n insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kv. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material. Take a parallel plate capacitor whose plates have an area of 2 m 2 and are separated by a distance of 1mm. The capacitor is charged to an initial voltage of 3 kv and then disconnected from the charging source. n insulating material is placed between the plates, completely filling the space, resulting in a decrease in the capacitors voltage to 1 kv. Determine the original and new capacitance, the charge on the capacitor, and the dielectric constant of the material. Given: 1 =3,000 V 2 =1,000 V = 2.00 m 2 d = 0.01 m Find: C =? C 0 =? Q =? κ =? Since we are dealing with the parallel-plate capacitor, the original capacitance can be found as 2 12 2 2 2.00 m C0 = ε0 = ( 8.85 10 C N m ) = 18 nf 3 d 1.00 10 m The dielectric constant and the new capacitance are C = κc = C = 318 nf = 54nF 1 0 0 2 The charge on the capacitor can be found to be ( )( ) 9 5 Q= C Δ V = F V = C 0 18 10 3000 5.4 10 7 8 2
How does an insulating dielectric material reduce electric fields by producing effective surface charge densities? Reorientation of polar molecules nduced polarization of non-polar molecules 17.1 Electric Current Whenever charges of like signs move in a given direction, a current is said to exist. Consider charges are moving perpendicularly to a surface of area. Definition: the current is the rate at which charge flows through this surface. Dielectric Breakdown: breaking of molecular bonds/ionization of molecules. 9 10 17.1 Electric Current - Definition 17.1 Electric Current - Units Given an amount of charge, ΔQ,, passing through the area in a time interval Δt,, the current is the ratio of the charge to the time interval. ΔQ = Δ t The S units of current is the ampere (). 1 = 1 C/s 1 of current is equivalent to 1 C of charge passing through the area in a time interval of 1 s. 11 12 3
17.1 Electric Current Remark 1 Currents may be carried by the motion of positive or negative charges. t is conventional to give the current the same direction as the flow of positive charge. 17.1 Electric Current Remark 2 n a metal conductor such as copper, the current is due to the motion of the electrons (negatively charged). The direction of the current in copper is thus opposite the direction of the electrons. - - - - v 13 14 17.1 Electric Current Remark 3 17.1 Electric Current Remark 4 n a beam of protons at a particle accelerator (such as RHC at Brookhaven national laboratory), the current is the same direction as the motion of the protons. n gases (plasmas) and electrolytes (e.g. Car batteries), the current is the flow of both positive and negative charges. t is common to refer to a moving charge as a mobile charge carrier. n a metal the charge carriers are electrons. n other conditions or materials, they may be positive or negative ions. 15 16 4
17.1 Electric Current Example Current in a light bulb The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. () the current in the light bulb. (B) the number of electrons that pass through the filament in 1 second. The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. () the current in the light bulb. ΔQ 1.67C = = = 0.835 Δt 2.00s 17 18 The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. (b) the number of electrons that pass through the filament in 1 second. The amount of charge that passes through the filament of a certain light bulb in 2.00 s is 1.67 c. Find. (b) the number of electrons that pass through the filament in 1 second. Reasoning: n 1 s, 0.835 C of charge passes the cross-sectional sectional area of the filament. This total charge per second is equal to the number of electrons, N, times the charge on a single electron. Solution: 19 ( ) N = N 1.60 10 C/ electron = 0.835C q 0.835C N = 19 1.60 10 C/ electron 18 N = 5.22 10 electrons 19 20 5
17.2 Current and Drift Speed Consider the current on a conductor of cross-sectional sectional area. q v d Δx = v d Δt 17.2 Current and Drift Speed (2) Volume of an element of length Δx is : V = Δx. Let n be the number of carriers per unit of volume. The total number of carriers in V is: n Δx. The charge in this volume is: ΔQ Q = (n Δx)q. Distance traveled at drift speed v d by carrier in time Δt: Δx = v d Δt. Hence: ΔQ Q = (n v d Δt)q. The current through the conductor: = ΔQ/ Δt t = n v d q. 21 22 17.2 Current and Drift Speed (3) 17.2 Current and Drift Speed - Example n an isolated conductor, charge carriers move randomly in all directions. When an external potential is applied across the conductor, it creates an electric field inside which produces a force on the electron. Electrons however still have quite a random path. s they travel through the material, electrons collide with other electrons, and nuclei, thereby losing or gaining energy. The work done by the field exceeds the loss by collisions. The electrons then tend to drift preferentially in one direction. Question: copper wire of cross-sectional sectional area 3.00x10-6 m 2 carries a current of 10.. ssuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The e density of copper is 8.95 g/cm 3. 23 24 6
Question: copper wire of cross-sectional area 3.00x10-6 m 2 carries a current of 10. ssuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cm 3. Reasoning: We know: = 3.00x10-6 m 2 = 10. ρ = 8.95 g/cm 3. q = 1.6 x 10-19 C. n = 6.02x10 23 atom/mol x 8.95 g/cm 3 x ( 63.5 g/mol) -1 n = 8.48 x 10 22 electrons/ cm 3. Question: copper wire of cross-sectional area 3.00x10-6 m 2 carries a current of 10. ssuming that each copper atom contributes one free electron to the metal, find the drift speed of the electron in this wire. The density of copper is 8.95 g/cm 3. ngredients: = 3.00x10-6 m 2 ; = 10.; ρ = 8.95 g/cm 3.; q = 1.6 x 10-19 C. n = 8.48 x 10 22 electrons/cm 3 = 8.48 x 10 28 electrons/m 3. 10.0 C/ s vd = = nq electrons m C m ( 28 3 )( 19 8.48 10 1.6 10 )( 3.00 10 6 2 ) 4 2.46 10 / = m s 25 26 17.2 Current and Drift Speed - Comments Mini-quiz Drift speeds are usually very small. Drift speed much smaller than the average speed between collisions. Electrons traveling at 2.46x10-4 m/s would would take 68 min to travel 1m. So why does light turn on so quickly when one flips a switch? The info travels at roughly 10 8 m/s Consider a wire has a long conical shape. How does the velocity of the electrons vary along the wire? Every portion of the wire carries the same current: as the cross sectional area decreases, the drift velocity must increase to carry the same value of current. This is dues to the electrical field lines being compressed into a smaller area, thereby increasing the strength of the electric field. 27 28 7
17.4 Resistance and Ohm s s Law - ntro 17.4 Definition of Resistance When a voltage (potential difference) is applied across the ends of a metallic conductor, the current is found to be proportional to the applied voltage. n situations where the proportionality is exact, one can write. Δ V = R The proportionality constant R is called resistance of the conductor. The resistance is defined as the ratio. R = 29 30 17.4 Resistance - Units 17.4 Ohm s s Law n S, resistance is expressed in volts per ampere. special name is given: ohms (Ω).( Example: if a potential difference of 10 V applied across a conductor produces a 0.2 current, then one concludes the conductors has a resistance of 10 V / 0.2 = 50 Ω. Resistance in a conductor arises because of collisions between electrons and fixed charges within the material. n many materials, including most metals, the resistance is constant over a wide range of applied voltages. This is a statement of Ohm s s law. Georg Simon Ohm (1787-1854) 31 32 8
Linear or Ohmic Material Non-Linear or Non-Ohmic Material Ohm s Law Δ V = R R understood to be independent of. Most metals, ceramics Semiconductors e.g. diodes 33 34 Definition: Example: Resistance of a Steam ron Resistor: a conductor that provides a specified resistance in an electric circuit. The symbol for a resistor in circuit diagrams. - V = R E ll household electric devices are required to have a specified resistance (as well as many other characteristics ). Consider that the plate of a certain steam iron states the iron carries a current of 7.40 when connected to a 120 V source. What is the resistance of the steam iron? 120V R = = = 16.2 Ω 7.40 35 36 9
17.5 Resistivity - ntro Electrons moving inside a conductor subject to an external potential constantly collide with atoms of the conductor. They lose energy and are repeated re-accelerated by the electric field produced by the external potential. The collision process is equivalent to an internal friction. This is the origin of a material s resistance. 17.5 Resistivity - Definition The resistance of an ohmic conductor is proportional to the its length, l,, and inversely proportional to the cross section area,,, of the conductor. l R = ρ The constant of proportionality ρ is called the resistivity of the material. 37 38 17.5 Resistivity - Remarks 17.5 Resistivity - Units Every material has a characteristic resistivity that depends on its electronic structure, and the temperature. Good conductors have low resistivity. nsulators have high resistivity. nalogy to the flow of water through a pipe. l R = ρ ρ = R l Resistance expressed in Ohms, Length in meter. rea are m 2, Resistivity thus has units of Ωm. 39 40 10
Resistivity of various materials (also see table 17.1) Material Resistivity (10-8 Ωm) Material Resistivity (10-8 Ωm) Silver 1.61 Bismuth 106.8 Copper 1.70 Plutonium 141.4 Gold 2.20 Graphite 1375 luminum 2.65 Germanium 4.6x10 7 Pure 3.5 Diamond 2.7x10 9 Silicon Calcium 3.91 Deionized 1.8x10 13 water Sodium 4.75 odine 1.3x10 15 Tungsten 5.3 Phosphorus 1x10 17 Brass 7.0 Quartz 1x10 21 Uranium 30.0 lumina 1x10 22 Mercury 98.4 Sulfur 2x10 23 41 Mini-quiz Why do old light bulbs give less light than when new? nswer: The filament of a light bulb, made of tungsten, is kept at high temperature when the light bulb is on. t tends to evaporate, i.e. to become thinner, thus decreasing in radius, and cross sectional area. ts resistance increases with time. The current going though the filament then decreases with time and so does its luminosity. Tungsten atoms evaporate off the filament and end up on the inner surface of the bulb. Over time, the glass becomes less transparent and therefore less luminous. 42 17.5 Resistivity - Example (a) Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321 m. = πr = π 0.321 10 m = 3.24 10 m Cross section: ( ) 2 2 3 7 2 Resistivity (Table): 1.5 x 10 6 Ωm. 6 Resistance/unit length: R ρ 1.5 10 Ωm = = = 4.6 7 2 l 3.24 10 m Ω m 17.5 Resistivity - Example (b) f a potential difference of 10.0 V is maintained across a 1.0-m length of the nichrome wire, what is the current? 10.0V = = = 2.2 R 4.6Ω 43 44 11
17.6 Temperature Variation of Resistance -ntro... next lecture The resistivity of a metal depends on many (environmental) factors. The most important factor is the temperature. For most metals, the resistivity increases with increasing temperature. The increased resistivity arises because of larger friction caused by the more violent motion of the atoms of the metal. 45 12