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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 concepts
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)
charges
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
atomic structure
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

electrostatic force coulomb s law
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)
practice coulombs law
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
practice coulomb s law
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
electric field e
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

electric field strength
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)
electric field strength1
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) 
practice electric field
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
practice electric field1
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
practice electric field2
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
electrostatic toys
Electrostatic “Toys”
  • Van de Graaff Generator
  • Wimshurst Machine
electrical potential difference voltage v
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.
practice potential difference
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
parallel plates
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)
practice parallel plates
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
example of parallel plates
Example of Parallel Plates
  • Lightning
    • Clouds - earth
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