ELECTRICITY & MAGNETISM NOTES

ELECTRICITY & MAGNETISM NOTES PHYSICS B4B BAKERSFIELD COLLEGE Rick Darke (Instructor)

CHARGE Electric charge is a fundamental property associated with 2 of the 3 subatomic particles making up most matter. 'What charge is' is unknown. 'How charge behaves' is well-known. e - p p n n e - e - p n electrons (negative charge) protons (positive charge) neutrons (no net charge) helium-4 atom ( 4 He)

CHARGE Charge is a scalar quantity that can be positive or negative. The standard unit of charge (S.I.) is the coulomb (C). Traditionally the symbols q and Q are used to denote charge. The smallest common unit of charge is the charge on an electron or proton: e - q electron = -1.6 x 10-19 C = -e p q proton = +1.6 x 10-19 C = +e n q neutron = 0 C The symbol e is used to represent the physical quantity fundamental charge (1.6 x 10-19 C).

QUARKS In 1963 Murray Gell-Mann and Georg Zweig proposed that hadrons (protons, neutrons, and other particles) are composed of 2 or 3 quarks. In 1979 Sheldon Glashow and Abdus Salam won the Nobel Prize in physics for work leading to the theory of how quarks interact with each other. Gell-Mann Zweig Glashow Salam

CHARGE OF QUARKS proton neutron u u u d d d u = up quark q u = +(2/3)e d = down quark q d = -(1/3)e

CHARGE MOBILITY Most materials can be classified as conductors, insulators, or semiconductors, depending upon how easy it is for charge to move through the material. conductor - a substance that allows charge to move through it with relative ease. examples: aluminum, gold, copper (shown) insulator - a substance through which charge move very reluctantly, or not at all. examples: polystyrene, rubber, glass (shown) semiconductor - a substance with electrical properties intermediate to those of conductors and insulators. examples: germanium, silicon (shown)

CONDUCTIVITIES OF MATERIALS class examples conductivity (Ω. m) -1 silver (Ag) 6.3 x 10 7 copper (Cu) 5.9 x 10 7 aluminum (Al) 3.6 x 10 7 carbon (C) 2.9 x 10 4 conductors semi- conductors insulators silicon (Si).016 germanium (Ge) 2.17 GaAs 1.0 x 10-6 glass ~10-12 nylon 6,6 ~10-12 quartz ~10-18 note: Most conductors become more conductive as their temperature increases, and most semiconductors become less conductive as their temperature increases.

LAW OF CHARGES Charges of the same sign are mutually repulsive, and charges of opposite sign are mutually attractive. q 1 F 21 F 12 q 2 _ + r F 21 q 1 q 2 F 12 + + r F 21 q 1 q 2 F 12 r note: The notation F 21 is taken to mean "the force due to charge q 2 and acting on charge q 1 ".

COULOMB'S LAW: QUALITATIVE The electric force between two point charges q 1 and q 2 is: (1) directed along the line joining the charges; (2) directly proportional to the product of the magnitudes of the charges; (3) inversely proportional to the distance between the charges squared; and (4) attractive if the charges have opposite signs and repulsive if they have like signs (Law of Charges). F 21 F 12 q 1 q 2 r note: Newton's Third Law insures that F 12 = - F 21 and that the two forces lie along the same line of action.

COULOMB'S LAW: QUANTITATIVE The electric force due to point charge q 1 acting on point charge q 2 is given by: q 1 q 2 r F 12 F 12 = k e q 1 q 2 n 12 /r 2 note: n 12 is a unit vector in the direction of r 2 - r 1 and k e is the Coulomb constant 9.0 x 10 9 Nm 2 /C 2.

USING COULOMB'S LAW Effective use of Coulomb's Law to compute the electric force (magnitude and direction) due to point charge q 1 acting on point charge q 2 : q 1 q 2 r F 12 (1) Compute the magnitude of the force using: F 12 = k e q 1 q 2 /r 2 (2) Use the Law of Charges to determine the direction of the force. (3) Express the combined magnitude and direction as the answer.

PRACTICE PROBLEM: 1-D COULOMB'S LAW Find the electric force on the -2.0 μc charge below due to the +4.0 μc charge to the left of it. +4.0 μc -2.0 μc 3.0 cm

PRACTICE PROBLEM: 2-D COULOMB'S LAW Find the electric force on the +1.0 μc charge located in the diagram below due to all other charges. y -3.0 μc 3.0 cm +1.0 μc +5.0 μc 4.0 cm x

H 2 0 MOLECULE CHARGE DISTRIBUTION + + + + + + H 104 o H O

PRACTICE PROBLEM: 2-D ELECTRIC FIELD Find the electric field at point P on the y-axis in the diagram below due to the two charges. y P 6.0 cm +1.0 μc -5.0 μc 8.0 cm x

PRACTICE PROBLEM: 1-D ELECTRIC FIELD Find all positions along the x-axis (other than infinity) where the electric field would be zero.. +4.0 mc -2.0 mc x = 0.0 m x = +1.0 m x

ELECTRIC MONOPOLE FIELD Electric field lines in the vicinity of a single point charge (electric monopole). The charge could be positive or negative, as either would show the same radial pattern. The dark lines in the photo are small pieces of thread suspended in oil, which align with the electric field produced by the single charged conductor at the center.

ELECTRIC DIPOLE FIELD Electric field lines in the vicinity of a point charge distribution where the charges have the same magnitude, but opposite sign (electric dipole). The dark lines in the photo are small pieces of thread suspended in oil, which align with the electric field produced by the two charges.

ELECTRIC FIELD Electric field lines in the vicinity of a point charge distribution where the charges have the same magnitude and sign. The dark lines in the photo are small pieces of thread suspended in oil, which align with the electric field produced by the two charges.

ELECTRIC FIELD: CHARGED CONDUCTORS Electric field pattern in the vicinity of two oppositely charged conductors. Note that all field lines are perpendicular to the conductors at their surfaces. The dark lines in the photo are small pieces of thread suspended in oil, which align with the electric field produced by the two charged conductors.

CHARGE ON AN ISOLATED CONDUCTOR Any excess charge on an isolated conductor resides entirely on its surface. The electric field is 0 everywhere inside the conductor. The electric field just outside a charged conductor is perpendicular to its surface (at the surface). The surface charge density is highest at sharp points (small radius of curvature of surface) on the surface.

PRACTICE PROBLEM: KINEMATICS A proton is released from rest at x = 0.0 cm in a region of uniform electric field. The magnitude of the electric field is 1500 N/C, and it is directed in the positive x-direction. Calculate the following quantities: (a) the force acting on the proton in the electric field; (b) the velocity of the proton when it reaches the position x = 5.0 cm; (c) the work done on the proton by the electric field for this displacement; (d) the change in the electric potential energy associated with the proton for this displacement. The mass of a proton is 1.67 x 10-27 kg.

USE OF k e and ε o The Coulomb constant k e (8.99 x 10 9 N.m 2 /C 2 ) has been used in some formulae in physics, while the permitivity of free space ε o (8.85 x 10-12 C 2 / N.m 2 ) has been used in others. The difference in usage is historical preference, where the two constants are related by: examples: ε o = 1/(4πk e ) k e is usually used in the relations: F 12 = k e q 1 q 2 /r 2 and V = k e q/r ε o is usually used in the relations: C = kaε o /d and E = σ/ε o

PRACTICE PROBLEM: CHARGED PLATES Two identical parallel rectangular plates with surface area.05 m 2 are placed a distance 1.0 cm apart. Charges of +5.0 nc and -5.0 nc are uniformly distributed over the surfaces of the plates. Calculate: (a) the magnitude of the surface charge density on the plates; and (b) the magnitude of the electric field that this charge distribution creates between the plates.

PARALLEL-PLATE CAPACITOR A parallel-plate capacitor consists of two parallel conducting plates with identical surface area (A) separated by a distance (d). The plates are charged equally and oppositely (+q and -q) by applying an electric potential difference (V) across the them. This creates uniform surface charge densities (+σ and -σ) on the two plates, producing an electric field (E) between the plates that is uniform between the plates and practically zero outside. The capacitor stores energy (U) in the process. + + + + + + + + + + + + + + A + + + + + + + + E +q -q useful relations: surface charge density: σ = q/a electric field: E = σ/ε o = V/d capacitance of capacitor: C = q/v = Aε o /d energy stored in capacitor: U = (1/2)CV 2 = (1/2)(1/C)q 2 + _ V

PRACTICE PROBLEM: CAPACITOR A parallel-plate capacitor has an area of 2.0 x 10-4 m 2 and a plate separation of 1.0 mm. The capacitor is connected to a 3.0-volt battery. Find the following quantities: (a) the capacitance of the capacitor; (b) the charge on the plates of the capacitor; (c) the charge density on the plates of the capacitor; (d) the magnitude of the electric field between the plates of the capacitor; and (e) the energy stored in the capacitor.

PRACTICE PROBLEM: CAPACITANCE A 24-volt source is connected to the parallel combination of capacitors shown below. Find the following quantities: (a) the charge on each capacitor; (b) the voltage drop across each capacitor; (c) the energy stored in each capacitor; and (d) the equivalent capacitance of the combination. 24 V +.003 F.006 F.012 F _

PRACTICE PROBLEM: CAPACITANCE A 9-volt source is connected to the parallel combination of capacitors shown below. Find the following quantities: (a) the charge on each capacitor; (b) the voltage drop across each capacitor; (c) the energy stored in each capacitor; and (d) the equivalent capacitance of the combination..08 F 9 V +.08 F _.12 F

PRACTICE PROBLEM: CAPACITANCE A 12-volt source is connected to the parallel combination of capacitors shown below. Find the following quantities: (a) the charge on each capacitor; (b) the voltage drop across each capacitor; (c) the energy stored in each capacitor; and (d) the equivalent capacitance of the combination..009 F 12 V +.006 F.012 F _

PRACTICE PROBLEM: CAPACITANCE Calculate the equivalent capacitance for the combination of capacitors shown below with respect to terminals a-b. 4 mf 1 mf a 6 mf 3 mf b 2 mf 8 mf

TABLE OF DIELECTRIC MATERIALS material κ (@ 20 C) E max (V/m) vacuum 1.0000 --------- dry air 1.0006 3 x 10 6 teflon 2.1 60 x 10 6 polystyrene 2.6 24 x 10 6 paper 3.7 16 x 10 6 pyrex 5.6 14 x 10 6 neoprene 6.7 12 x 10 6 mica 7 150 x 10 6 water 80 --------- SrTiO 3 233 15 x 10 6

EXERCISE IN CAPACITOR ENERGETICS ACTION 1: Start with a 2.0-F air-spaced capacitor with uncharged plates (no voltage dropped across them). Use a variable voltage source to drop 1.0 volt across the plates of the capacitor. ACTION 1 C = 2.0 F V = 0.0 V q = 0.0 C U = 0.0 J C = V = 1.0 V q = U = + _ V + _ V

EXERCISE IN CAPACITOR ENERGETICS ACTION 2: Insert a slab of dielectric material with a dielectric constant of 3.0 between the plates of the capacitor. Do not disconnect the voltage source or change the voltage drop across the plates. ACTION 2 C = 2.0 F V = 1.0 V q = 2.0 C U = 1.0 J κ = 3.0 C = V = q = U = + _ V + _ V

EXERCISE IN CAPACITOR ENERGETICS ACTION 3: Use the voltage source to increase the voltage drop across the plates of the capacitor to 5.0 V. Do not change anything else. ACTION 3 κ = 3.0 C = 6.0 F V = 1.0 V q = 6.0 C U = 3.0 J κ = 3.0 C = V = 5.0 V q = U = + _ + _ V V

EXERCISE IN CAPACITOR ENERGETICS ACTION 4: Disconnect the leads of the voltage source from the plates of the capacitor and remove the source. Do not change anything else. ACTION 4 κ = 3.0 C = 6.0 F V = 5.0 V q = 30 C U = 75 J κ = 3.0 C = V = q = U = + _ V

EXERCISE IN CAPACITOR ENERGETICS ACTION 5: Remove the slab of dielectric material from between the plates of the capacitor. Do not change anything else. ACTION 5 κ = 3.0 C = 6.0 F V = 5.0 V q = 30 C U = 75 J C = V = q = U =

RESISTIVITIES OF MATERIALS material ρ o @ 20 C (Ω.m) α ( C -1 ) silver 1.6 x 10-8.0038 copper 1.7 x 10-8.0039 aluminum 2.8 x 10-8.0039 tungsten 5.6 x 10-8.0045 nichrome 1.5 x 10-6.0004 carbon 3.5 x 10-5 -.0005 silicon 640 -.0750 glass ~10 12 quartz ~10 18 note: To obtain the resistivity (ρ) of a material in the table above at a specific celsius temperature (T), use: ρ = ρ o (1 + α(t - T o ), where T o = 20 C

CYLINDRICAL RESISTORS R = IV R = ρl/a P = IV = I 2 R = V 2 /R E = V/L A E L L = length of cylindrical resistor (m) A = cross-sectional area of resistor (m 2 ) ρ = resistivity of conducting material (Ω.m) T = temperature of resistive material ( C) R = resistance of resistor (Ω) V = voltage dropped across resistor (V) I = current through resistor (A) E = uniform electric field inside resistor (V/m) P = power dissipated by resistor (W) + _ V

PRACTICE PROBLEM: RESISTANCE A nichrome wire with a diameter of 0.50 mm and length of 5.0 meters is used as a heating element. In use the element has a temperature of 1500 C when 120 volts is dropped across the ends. Find: (a) the resistivity of nichrome at this temperature; (b) the resistance of the element; (c) the current through the element; and (d) the power dissipated by the element.

PRACTICE PROBLEM: RESISTANCE Suppose that a platinum thermistor is used to interpret the temperature of the environment in which it is immersed. It has a resistance of 50.0 Ω at 20 C. When it is immersed in a vessel containing melting indium the resistance of the thermistor is found to increase to 130.0 W. Use this information to find the melting point of indium.

PRACTICE PROBLEM: RESISTANCE An immersion heater whose resistance in operation is 24 ohms is placed in a mug with 400 grams of water initially at a temperature of 20 C. If the heater is connected to a 120-volt source, how long will it be until the water comes to a boil? Assume the mug is a good thermal insulator and does not absorb heat during the process.

KIRCHHOFF'S CIRCUIT LAWS Kirchhoff's Current Law (KCL): The sum of all currents directed into a junction is equal to the sum of all currents directed out of the junction: Σ I in = Σ I out 15 Ω J 18 Ω K 20 Ω 2.0 A 1.0 A 0.6 A 60 V 1.0 0.4 + _ A 30 Ω A 30 Ω note: At junction J, Σ I in = 2.0 A and Σ I out = 1.0 A + 1.0 A = 2.0 A, and at junction K, Σ I in = 1.0 A and Σ I out = 0.4 A + 0.6 A = 1.0 A

KIRCHHOFF'S CIRCUIT LAWS Kirchhoff's Voltage Law (KVL): The sum of all voltage gains around any loop in a circuit is equal to zero: Σ ΔV loop = 0 15 Ω J 18 Ω K 20 Ω 2.0 A 1.0 A 0.6 A 60 V 1.0 0.4 + _ A 30 Ω A 30 Ω note: Around the outermost mesh, Σ ΔV loop = 60 V - (15 Ω)(2.0 A) - (18 Ω)(1.0 A) - (20 Ω)(0.6 A) = 0 V