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Learn about the origin of charge, conductors vs. insulators, Coulomb’s Law, and electric field concepts in this comprehensive lecture on electric fields and forces. Explore charge properties, units, conductors, insulators, and the practical application of Coulomb’s Law.
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Lecture 01: Electric Fields & Forces • Origin of Charge • Conductors and Insulators • Coulomb’s Law • The Electric Field
Charge • Charge is an intrinsic property of matter. • Units: Coulombs (C) • Two types: • Positive Charge: Protons (Q = +1.6 x 10-19 C) • Negative Charge: Electrons (Q = -1.6 x 10-19 C) • Opposites Attract - Likes Repel • In general, atoms are neutral: • Negatively charged electrons “orbit”… • Atomic Radius approximately 10-10 m • …Positively charged central nucleus. • Nucleus Radius approximately 10-15 m + - + -
Conductors & Insulators • Perfect Conductors: • Electrons free to move throughout the conductor. • Perfect Insulators: • Electrons fixed to atoms throughout the insulator. • Most materials are somewhere in-between.
Coulomb’s Law • Gives the magnitude of the force between two charges: • Coulomb’s constant k = 9 x 109 N-m2/C2 • R is the distance between the two charges. • Remember: • Opposite charges ATTRACT • Like charges REPEL
Summary of Concepts • Charge: a property of matter • Conductors vs. Insulators • Coulomb’s Law:a formula to calculate the electric force between two charges
Example • Given: two positive charges and a negative charge. • What is the magnitude and direction of the force on the middle charge due to the other two? Q2 = -2.5C Q1 = +1.5C Q3 = +3.5C - + + d1 = 2.0 m d2 = 1.5 m
Example • First, the direction of the force: • Both positive charges attract the middle negative charge. • However, since the charge on the right has a greater magnitude and is closer, the force toward the right will be greater. • The NET force is to the RIGHT. - + +
Example • Second, the magnitude of the force: • We need to use Coulomb’s Law for each charge: Q2 = -2.5C Q1 = +1.5C Q3 = +3.5C - + + d1 = 2.0 m d2 = 1.5 m
Example • Finally, subtract the two forces since they are in opposite directions: • The NET force is: 26.6 x 10-3 N to the right Q2 = -2.5C Q1 = +1.5C Q3 = +3.5C - + + d1 = 2.0 m d2 = 1.5 m
Example • First, the direction of the force: • Both positive charges attract the middle negative charge. • However, since the charge on the right has a greater magnitude and is closer, the force toward the right will be greater. • The NET force is to the RIGHT. - + +
The Electric Field • A charged particle creates an Electric Field. • Electric Field Lines are used to represent fields. • Field lines point in the direction a positive charge would accelerate. • Density of field lines is proportional to the strength of the field. • The Electric Field around a single point charge is:
The Electric Field • Fields lines point away form positive charges. • Fields lines point toward negative charges • The force on a charge in an Electric Field is: - + q E
Conductors • Electrons are free to move in a Conductor. • The Electric Field in a conductor is Zero. • Electric Field Lines are always Perpendicular to the surface of a conductor.
Summary of Concepts • Coulomb’s Law:a formula to calculate the electric force between charges. • The Electric Field: created bycharges • Conductors • Electrons are free to move • E = 0 inside • Field Lines
Example • Given: two positive charges and a negative charge on three corners of a square with side length 3.8 m. • What is the magnitude and direction of the electric field at the 4th corner of the square? + Q1 = +2 C Q2 = -2 C Q3 = +2 C - +
Example • First the direction of the Electric Field: • Since electric fields point away from positive charges and toward negative charges, the three vectors shown give the field from each individual point charge at the 4th corner. • Since the negative charge is farthest away, its electric field is the weakest, and the NET electric field will therefore be UP and to the RIGHT. + Q1 = +2 C Q2 = -2 C Q3 = +2 C - +
Example • Second, the magnitude of the field. We will use: • We need to calculate the field from each of the three charges: 1247 N/C 623 N/C 1247 N/C
Example • Finally, add the fields as VECTORS by adding the corresponding components: • The NET force is: + Q1 = +2 C Q2 = -2 C Q3 = +2 C - + x-component: 1247 - 623cos45 = +806 N/C y-component: 1247 - 623sin45 = +806 N/C 1140 N/C up and to the right