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Lecture 01: Electric Fields & Forces

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 )

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Lecture 01: Electric Fields & Forces

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  1. Lecture 01: Electric Fields & Forces • Origin of Charge • Conductors and Insulators • Coulomb’s Law • The Electric Field

  2. 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 + - + -

  3. 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.

  4. 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

  5. Summary of Concepts • Charge: a property of matter • Conductors vs. Insulators • Coulomb’s Law:a formula to calculate the electric force between two charges

  6. 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 = -2.5C Q1 = +1.5C Q3 = +3.5C - + + d1 = 2.0 m d2 = 1.5 m

  7. 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. - + +

  8. Example • Second, the magnitude of the force: • We need to use Coulomb’s Law for each charge: Q2 = -2.5C Q1 = +1.5C Q3 = +3.5C - + + d1 = 2.0 m d2 = 1.5 m

  9. 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 = -2.5C Q1 = +1.5C Q3 = +3.5C - + + d1 = 2.0 m d2 = 1.5 m

  10. 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. - + +

  11. 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:

  12. 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

  13. 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.

  14. 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 

  15. 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 - +

  16. 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 - +

  17. 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

  18. 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

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