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Published byHarold Harvey Modified over 9 years ago
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Charges
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Electrons P+ n nucleus e-
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Source of Electrons(Force)
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Matter : Weight - Space Solid Liquid Gas Plasma
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Atoms P n nucleus e- valence shell
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Atoms Atomic Number – Number of Protons Atomic Weight – Number of Protons plus Neutrons Number of Protons EQUALS Electrons in a neutral atom
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Ionization The process where an atom gains or loses an electron. Na + Cl - Na – Positive Ion Cl – Negative Ion
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Charges (q) Like Charges (++ - -) q- q+ Repel
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Charges (q) Un-Like Charges (+ -) q+q- Attract
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Coulomb’s Law Coulomb (C) The quantity of charge on an abject. The quantity of charge on an abject. 1C = 6.25 x 10 18 e -
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q- Coulomb’s Law
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q1q1 q2q2 What does the force depend on?
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q1q1 q2q2 Coulomb’s Law 1.The charge on q 1 and q 2. 2.The distance (d). d
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q- dd2d2
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q1q1 q2q2 Mathematical Model for Coulomb’s Law Force (F) d F ∞ q1q2 F ∞ 1/d 2
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q1q1 q2q2 d F = q1q2q1q2q1q2q1q2 d2d2d2d2 K K = 9.0 X 10 9 N·m 2 /C 2
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q1q1 q2q2 d=3cm d = 3cm = 3x10 -2 m K = 9.0 X 10 9 N·m 2 /C 2 q 1 = +6μC = +6x10 -6 C q 2 = +3μC = +6x10 -6 C What is the force on q 2 ?
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F = (3X10 -2 m) 2 9X10 9 N·m 2 /C 2 (6X10 -6 C)(6X10 -6 C) q1q2q1q2q1q2q1q2 d2d2d2d2K F = 1.8X10 2 N q1q1 q2q2
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Electric Fields Think of gravity or gravitational field to understand electric fields.
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Electric Fields vs Gravity g
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Electric Fields E neutral neutral ++++++++++ - - - - - - - - - -
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Electric Fields E + + ++++++++++ - - - - - - - - - -
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Electric Fields E - - ++++++++++ - - - - - - - - - -
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Electric Fields A force field due to electric charges. E A force on a charge due to another chage.
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Electric Field: Point Charge q E = kq r2r2r2r2 r
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Force on a charge ‘q’ in an Electric Field Q q E = Fq
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Fq NewtonCoulomb NC
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Energy and Work
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+ - It takes work to move the “-” charge.
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Electric Potential (Energy) Voltage + q’ q’ ΔV = Work on q’ q’ JouleCoulomb = = volt
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Electric Potential (Energy) Voltage + q’ q’ ΔV = V B - V A VAVAVAVA VBVBVBVB “Potential Difference” Voltage
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Electric Potential (Energy) Voltage ΔV = Fdq Eqdq E = Fq
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Electric Potential (Energy) Voltage ΔV = Ed
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Example Two parallel plates are given opposite charges. A voltmeter measures the electric potental differnece to be 60V. The plates are 3.ocm apart. What is the magnitude of the electric field between them.
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- - - - - - - - - - ++++++++++ E { ΔVΔV { d ΔV=60.0VD=0.030m
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ΔV = Ed = E ΔVdΔVdΔVdΔVd 60V 0.03m
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2000J/(m·C)= E 2000 N/C = E
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Current Flow of charge! Charge - Coulomb(C) 6.24 X 10 18 electrons eeeeee ee e e eeeeee
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Leyden Jar Conductors
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+++ + + + - - - --- --- - - - + ++ +++ - - -
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Capacitor A capacitor is a passive electronic component that stores energy in the form of an electrostatic field.
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Capacitor Metal Plates Insulating Material Dielectric
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Capacitor - + e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e-
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Capacitor-------- + +++ + +++
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Copper Zinc -
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Current eeeeee ee e e eeeeee Number of electrons passing a point in a second.
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Current Symbol - I Unit – Ampere (Amps) Amperes - A
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Force eeeeee ee e e eeeeee Voltage Potential Difference Electromotive Force (EMF)
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Voltage Symbol - V Unit – Volt Volt - V
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Resistance Opposition to Current Flow High Resistance Insulator
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Resistance Opposition to Current Flow Low Resistance Conductor
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Resistance Opposition to Current Flow High and Low Resistance SemiConductor
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Resistance Symbol - R Unit – Ohm Ohm - Ω
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Putting it all Together Load Source Path
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Homework: 1-1 Worksheet Due: 9/9/02 Test: 9/9/02
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