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Electrostatics

Electrostatics. Overview. Electrostatics - Concepts. Definition Electrostatics = Electron + stationary “The study of electric charges , forces and fields” Symbol for charge Letter “q” or “Q” Units of charge Coulombs (C) - standard

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Electrostatics

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  1. Electrostatics Overview

  2. Electrostatics - Concepts • Definition • Electrostatics = Electron + stationary • “The study of electric charges, forces and fields” • Symbol for charge • Letter “q” or “Q” • Units of charge • Coulombs (C) - standard • 1 C = the charge on 6.25 x 1018 electrons, therefore • 1 electron has a charge of 1.6 x 10-19 C • microCoulombs (µC) - common • X 10-6 (ex. 3 x 10-6 C = 3 µC)

  3. Charges • Charges can be either • Negative (-) • an excess of electrons • Positive (+) • a deficiency of electrons • “Fundamental” or “elementary” charge • Electron = -1.6 x 10-19C • “Quantized” – integer numbers

  4. Atomic Structure • Electrons are free to move, but not protons or neutrons • Positive and negative particles, when brought close together are subject to the following “BIG RULE”: • Like (+/+ or -/-) charges repel • Unlike (+/-) charges attract Overall – charge is conserved

  5. Electrostatic Force - Coulomb’s Law • Two charged particles q1 & q2 interact with each other with a force that is: • Proportional to the product of their charges, and • Inversely proportional to the square of their separation distance (d) • where • k = electrostatic constant • k = 9 x 109 N.m2/C2 • (Cf Newton’s Law of Universal Gravitation)

  6. Practice – Coulombs’ Law • What is the magnitude of the e-static force between an electron and proton of hydrogen, when they are separated 5.3 x 10-11 m? (qe = -1.6 x 10-19 C, qp = 1.6 x 10-19 C) • Solve Coulomb’s Law for F • F = k(q1q2)/d2 • F = 9 x 109 x (-1.6 x 10-19 C) x (1.6 x 10-19 C) (5.3 x 10-11)2 • F = 8.2 x 10-8 N

  7. Practice – Coulomb’s Law • Two e-static charges of 60 x 10-6C and 50 x 10-6C exert a force of 175N on each other. What is their separation distance? • Solve Coulomb’s Law for d: • F = k(q1q2)/d2 • 175 = 9 x 109 x (60 x 10-6)(50 x 10-6) d2 • d2 = 1.54 x 10-1 • d = 3.93 x 10-1 m

  8. Some results of E/static charges

  9. Electric Field (E) • An electric field is a region in space in which an electric charge experiences a force. • Surrounding electric charges are lines of force which form an electric field. These lines of force flow out of a positive charge and into a negative charge, as shown. • Represented by symbol E • E is a vector! E

  10. Electric Field Strength? • The Electric Field strength can be determined by 2 methods: • Method 1: The strength of this electric field (E) at a point is equal to the force felt by a small positive test charge (q0) being placed at that point. • E = F/q0 Newtons/Coulomb (N/C)

  11. Electric Field Strength? • Method 2: the electric field strength at a point P, d units distant from the charge of interest q is equal to the product of the charge and the electrostatic constant (k) divided by the square of the separation distance, or • E = kq/d2 (N/C) 

  12. Practice – Electric Field • A force of 2.1 N is exerted on a 9.2e-4 C test charge when it is placed in an electric field created by a 7.5 C charge. If the force is pushing it West, • determine the electric field (E) at that point. • Solve E = F/q0 for E • E = 2.1/9.2e-4 • E = 2,282.6 N/C

  13. Practice – Electric Field • If a positive test charge of 3.7e-6 C is put in the same place in the electric field as the original test charge in the last example, • determine what force (F) will be exerted on it. • Solve E = F/qo for F • 2282.6 = F/3.7e-6 • F = 8.44e-3 N

  14. Practice – Electric Field • A tiny metal ball has a charge of –3.0e-6 C. • determine the magnitude and direction of the field at a point, P, 30cm away? • Solve E = kq/d2 for E • E = (9e9)(-3e-6)/0.32 • E = -300,000 N/C

  15. Electrostatic “Toys” • Van de Graaff Generator • Wimshurst Machine

  16. Electrical Potential Difference (Voltage V) • Definition • The amount of work done (change in electrical potential energy) in moving a +ve test charge (q0)from one location to another in an electrical field (E). • V = W/q0 in joules/coulomb = volts • If the charge is moving against the field, then the work done is negative, and vice versa.

  17. Practice – Potential Difference • Determine the electric potential difference (or voltage V) of a 3.4 C charged object (q) that gains 2.6 x 103 J as it moves through an electric field. • Solve V =W/q for V • V = (2.6x 103)/3.4C • V = 7.6 x 102V

  18. Parallel Plates • If an electric field (E) exists between 2 parallel plates, then the field is uniform except at the edges. • The electric field depends on the potential difference (V) across the plates and the plate separation (d), as follows: • E = V/d, • units of volt/meter (equivalent to N/C)

  19. Practice – Parallel Plates • If we have two parallel plates that are 16.0mm apart and want a uniform field of 800 N/C between these plates, determine the voltage we must apply to the plates. • Solve E = V/d for V • V = (800N/C) (0.0160m) • V = 12.8 V

  20. Example of Parallel Plates • Lightning • Clouds - earth

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