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Electricity and Magnetism

Electricity and Magnetism. Unit III. I Electrostatics. The study of electric charges at rest and their electric fields and potentials Charges at rest if there is no net transfer of charge Microstructure of Matter The smallest unit of matter is an atom

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Electricity and Magnetism

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  1. Electricity and Magnetism Unit III

  2. I Electrostatics • The study of electric charges at rest and their electric fields and potentials • Charges at rest if there is no net transfer of charge • Microstructure of Matter • The smallest unit of matter is an atom • Atoms are made up of protons, neutrons and electrons • The electron is the fundamental negatively charged particle of matter • The proton is the fundamental positively charged particle of matter

  3. The elementary charge (e) equals • The charge on an electron (-e) • The charge on a proton (+e) • Because all atoms are electrically neutral, atoms contain equal number of protons and neutrons • Neutrons have no charge

  4. B. Charged Objects • Protons and neutrons cannot be removed from an atom • Electrically charged objects are formed when neutral objects gain or lose electrons • When an atom becomes a charged particle, it is called an ion • Excess electrons creates negative ions • Loss of electrons creates positive ions • Objects with the same sign of charge are repelled • Objects with opposite signs are attracted

  5. C. Transfer of Charge • If a system only contains neutral objects, the total net charge is zero • If the objects are rubbed together, electrons are transferred • The system as a whole remains neutral

  6. 0e R -8e S +6e T Example 3 spheres have initial charges • When R and S touch, excess electrons (-8e) move to neutral (0e) • When separated each sphere “splits” the excess • (-8e + 0e) ÷ 2 spheres • Each sphere becomes -4e • When S and T touch, excess electrons (-4e) moves to (+6e) • When separated each sphere “splits” the excess • (-4e + 6e) ÷ 2 spheres • Each sphere becomes +1e • TOTALS R = -4e S = +1e T = +1e

  7. D. Quantity of Charge • SI unit of charge is a coulomb (C) • 1C = 6.25 x 1018 elementary charges (electrons) • The charge on an electron (-e) = 1.6 x 10-19C • The net charge on a charged object is always a multiple of the charge on an electron

  8. E. Coulomb’s Law • The electrostatic force between 2 charges is • Directly related to the product of the charges • Inversely related to the square of the distance between them • F…electrostatic force (newtons) • q…charge (coulombs) • r…distance between charges (meters) • k…electrostatic constant (8.99 x 109 N m2/C2) kq1q2 F = r2

  9. II. Electric Fields • Region around a charged particle • Exerts a force on other charged particles • Represented by field lines • Direction a positive charge(+) would move • If lines are curved, charge would move tangent to the point on the field line • Begin (+) and end (-) • Never cross • Energy varies inversely squared with distance

  10. E = F q • Strength Where E = electric field strength (N/C) F = force the charge experiences (N) q = the charge (C)

  11. E = F q What is the magnitude of the electric field at a point in a field where an electron experiences a 1.0 N force? E = 6.3 x 1018 N/C E = 1.0 N 1.60 x 10-19C

  12. V = W q A. Potential Difference • Work done as charge is moved against a field • Where W is work in joules • q is charge in coulombs • V is potential difference • Joules/coulomb is the derived unit volt • 1 J/C = 1 volt • If the charge is an elementary charge (ex. electron) thepotential difference is an electron volt (eV) • 1 eV = 1.60 x 10-19J

  13. Moving a point charge of 3.2 x 10-19 coulombs between two points in an electric field requires 4.8 x 10-18 joules. What is the potential difference between these two points? V = W q V = 4.8 x 10-18 J 3.2 x 10-19C V = 15 V

  14. III Electric Current A. Electric current • The rate that a charge passes a point in a circuit • Represented using the symbol, I • Unit is an ampere (A) • Formula I = q t • I is current in amperes • q = charge in coulombs • t = time in seconds • Current is measured using an ammeter

  15. B. Electric circuit • The closed path that a charged particle moves along • Potential difference is also needed for an electric current • Supplied by • Cell • Battery • Measured using a voltmeter • Electron flow determines the direction of current • A switch is used to make, break or change the connections in a circuit

  16. C. Resistance • Electrical resistance (R) is the opposition of electron flow • Ratio of potential difference to current flow • R = V I • Where • V = potential difference in volts • I = current in amperes • R = resistance in volts per ampere • Ohm (Ω)represents 1 volt per ampere • Resistance is affected by temperature

  17. A current of 0.10 amperes flow through a lamp connected to a 12.0-volt source. What is the resistance of the lamp? I = 0.10 A V = 12.0 V R = V I R = 12.0 V 0.10 A R = 120 Ω

  18. Factors Affecting Resistance Length of Wire • Increasing length, increases the collisions of electrons with atoms in the wire • Directly related Thickness of Wire • Increasing thickness creates more spaces for electrons to travel through, decreasing the resistance • Inverse relationship with cross-sectional area Resistivity • Characteristic of the material • Good conductors have low resistivities • As temperature increases, resistivity increases • Values found in reference tables

  19. Resistance of a Wire R = ρL A Where • R is resistance in ohms (Ω) • ρ is the resistivity in ohm-meters (Ω-m) • L is length in meters (m) • A is cross-sectional area in square meters (m2)

  20. Determine the resistance of a 4.00-meter piece of copper wire that has a diameter of 2.00 mm at 20°C. ρ copper = 1.72 x 10-2 L = 4.00 m Diameter = 0.002 m Area = πr2 Area = π(.001)2 Area = 3.14 x 10-6 m2 R = (1.72 x 10-2 x 4.00) 3.14 x 10-6 R =2.19 x 10-2Ω

  21. Resistor • Device to have a definite amount of resistance • Used to limit current flow • Provide a potential drop • Symbol

  22. Electric Power Power (P) = Work/time Units Work,,,joules Time sec Power watts Electrical power is the product of VI Since V=IR then P = I2R

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