Physics 202, Exam 1 Review Logistics Topics: Electrostatics + Capacitors (Chapters 21-24) Point charges: electric force, field, potential energy, and potential Distributions: electric field, electric potential. Interaction of point charges with continuous distributions. Conductors: charge distribution, electric field, electric potential Capacitors: compute capacitance, energy stored in capacitor
Exam 1 Logistics Exam time: Wednesday, February 17, 5:30-7PM Rooms: 2103 Chamberlain and 2241 Chamberlain 2103: 304, 321, 310, 322, 305, 330, 324, 326, 307, 325, 301 2241: 302, 303, 327, 308, 329, 309, 323 Bring: Pen/pencil Calculator (no programming functionality) 1 single-sided formula sheet, self-prepared (no photocopying)
Exam 1: Electrostatics Topics, Mechanics Topics: Coulomb Forces, Potential Energy Electric Field and Potential of Point Charges and Distributions Motion of charged particles in electric fields Electric field lines and equipotentials Conductors in electrostatic equilibrium Capacitance, Capacitors in circuits and dielectrics Mechanics -- not the main focus, but you should know: Kinematics of uniformly accelerated particles Newton s Laws: statics and dynamics Anything on homework is fair game for this midterm or the final exam (e.g. circular orbits, springs) Math -- you will not be expected to do nontrivial integrals. You should be able to do integrals which require simple substitutions.
Topics: Point Charges (I) 2 charges: force on q 2 by q 1 F 12 F 12 = k q 1q 2 r 2 r 12 = F 21 F 21 >2 charges: force on charge i F = F + F + F +... i 1i 2i 3i ˆr 12 F 21 F 12 principle of linear superposition
Electric Field: Electric Potential: Topics: Point Charges (II) Field concept: electric field and electric potential V = F = q E E = i E i q V i = k i i i r i = k i q i r i 2 ˆr i V B V A = Relation between force and field: ΔU = qδv B A r = vector from source to observation point E d l E = V
Gauss s Law Karl Friedrich Gauss Net electric flux through any closed surface ( Gaussian surface ) equals the total charge enclosed inside the closed surface divided by the permittivity of free space. electric flux q encl : all charges enclosed regardless of positions Gaussian Surface Φ E = Eid A = q encl ε 0 ε 0 : permittivity constant k = 1 4πε 0
Using Gauss Law Choose a closed (Gaussian) surface such that the surface integral is trivial. Use symmetry arguments: 1. Direction. Choose a Gaussian surface such that E is clearly either parallel or perpendicular to each piece of surface 2. Magnitude. Choose a surface such that E is known to have the same value at all points on the surface Then: EidA = E da = E da = EA = q encl ε 0 Given q encl, can solve for E (at surface), and vice versa
Continuous Charge Distributions (I) Method 1: high degree of symmetry Use Gauss s Law to obtain E. Integrate to get V. E d A = q encl ε 0 V(r) = r ref E d r Examples: spherical symmetry, cylindrical symmetry, planar symmetry Conductors (surface charge density only) and insulators
Basic Symmetries Φ E = Eid A = q encl ε 0 Use it to obtain E field for highly symmetric charge distributions. spherical cylindrical planar (point charge, uniform sphere, spherical shell, ) (infinite uniform line of charge or cylinder ) (infinite uniform sheet of charge, ) Method: evaluate flux over carefully chosen Gaussian surface
Continuous Charge Distributions (II) 1. Direct calculation of E field: integrate to get V. d E = k dq r 2 ˆr E = d E 2. Direct calculation of V: take derivatives to get E. dv = k dq r V = Examples: uniformly charged ring, disk (on-axis), finite line charge. dv
Conductors and Capacitors (I) Main feature of conductors: Electrostatic equilibrium: E = 0 inside conductor E = σ ε 0 outside, perp to surface (only surface charges) equipotentials Application: Capacitors
Capacitors Definition: Q = CV Computing capacitance: Parallel plate (also know: spherical, coaxial) Capacitors in circuits: C = ε 0 A Charging/discharging Series and Parallel combinations d U = 1 2C Q2 = 1 2 CV 2 1 C S = i 1 C i C P = i C i
Capacitors: Summary Definition: C Q ΔV Capacitance depends on geometry: A + + + + d - - - - - r +Q a b -Q L -Q +Q a b Parallel Plates C = ε o A d Cylindrical C = 2πε L o ln b a C has units of Farads or F (1F = 1C/V) ε o has units of F/m Spherical C = 4πε o ab b a
Dielectrics Empirical observation: Inserting a non-conducting material (dielectric) between the plates of a capacitor changes the VALUE of the capacitance. Definition: The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it: κ = C C 0 κ permittivity: ε κε 0 κ values are always > 1 (e.g., glass = 5.6; water = 80) Dielectrics INCREASE the capacitance of a capacitor More energy can be stored on a capacitor at fixed voltage: E = E 0 κ U ʹ = CV 2 2 = κc 0 V 2 2 = κu
A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q, A. the capacitance becomes (1/2)V. B. the capacitance becomes 2C. C. the potential changes to (1/2)V. D. the potential changes to 2V. E. the potential does not change.
A capacitor of capacitance C holds a charge Q when the potential difference across the plates is V. If the charge Q on the plates is doubled to 2Q, A. the capacitance becomes (1/2)V. B. the capacitance becomes 2C. C. the potential changes to (1/2)V. D. the potential changes to 2V. E. the potential does not change.
If C 1 < C 2 < C 3 < C 4 for the combination of capacitors shown, the equivalent capacitance is A.less than C 1. B.more than C 4. C.between C 1 and C 4.
If C 1 < C 2 < C 3 < C 4 for the combination of capacitors shown, the equivalent capacitance is A.less than C 1. B.more than C 4. C.between C 1 and C 4.
If C 1 < C 2 < C 3 < C 4 for the combination of capacitors shown, the equivalent capacitance is A.less than C 1. B.more than C 4. C.between C 1 and C 4.
If C 1 < C 2 < C 3 < C 4 for the combination of capacitors shown, the equivalent capacitance is A.less than C 1. B.more than C 4. C.between C 1 and C 4.
Two identical capacitors A and B are connected across a battery, as shown. If mica (κ = 5.4) is inserted in B, A. both capacitors will retain the same charge. B. B will have the larger charge. C. A will have the larger charge. D. the potential difference across B will increase. E. the potential difference across A will increase.
Two identical capacitors A and B are connected across a battery, as shown. If mica (κ = 5.4) is inserted in B, A. both capacitors will retain the same charge. B. B will have the larger charge. C. A will have the larger charge. D. the potential difference across B will increase. E. the potential difference across A will increase.
Example A capacitor is charged with a battery to a charge of Q=3.30 10-7 C. The area of the plates is 100 cm 2 =0.01 m 2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0 10 6 V/m. 1) What is the energy stored in the capacitor with the dielectric present?
Example A capacitor is charged with a battery to a charge of Q=3.30 10-7 C. The area of the plates is 100 cm 2 =0.01 m 2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0 10 6 V/m. 1) What is the energy stored in the capacitor with the dielectric present? U = ½QV = ½Q(Ed) = ½(3.03 10-7 )(3.03 10-7 )(0.01) U = 1.52 10-3 J
Example A capacitor is charged with a battery to a charge of Q=3.30 10-7 C. The area of the plates is 100 cm 2 =0.01 m 2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0 10 6 V/m. The battery is disconnected and then the dielectric is removed, so the capacitor plates are separated by air. What is the energy stored in the capacitor after the dielectric has been removed?
Example A capacitor is charged with a battery to a charge of Q=3.30 10-7 C. The area of the plates is 100 cm 2 =0.01 m 2, and the distance between the plates is 1 cm=0.01 m. With dielectric inserted into the capacitor, the electric field is found to be 1.0 10 6 V/m. The battery is disconnected and then the dielectric is removed, so the capacitor plates are separated by air. What is the energy stored in the capacitor after the dielectric has been removed? Charge is same: Q=3.03 10-7 C Electric field without dielectric: E 0 = σ ε 0 = Q ε 0 A = 3.03 10 7 ( 8.85 10 12 ) 0.01 ( ) == 3.42 106 V / m κ = E E 0 = 3.42 106 1.0 10 6 = 3.42 U 0 = κu = (3.42)(1.52 10 3 ) = 5.2 10 3 J.