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Electric Fields

Electric Fields. Applied Physics and Chemistry Electricity Lecture 3 (I think). Electric Field. Definition: Michael Faraday (1791-1867) Electric field exists around any charged object Area where a test charge would be given a force by the charged object

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Electric Fields

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  1. Electric Fields Applied Physics and Chemistry Electricity Lecture 3 (I think)

  2. Electric Field • Definition: • Michael Faraday (1791-1867) • Electric field exists around any charged object • Area where a test charge would be given a force by the charged object • Test charge is a positive charge of small magnitude • Magnitude of field is a measure of the force exerted by the charged object • Area of field indicated by field lines • Show the path that would be taken by the test charge • Note: electric field intensity is a VECTOR!

  3. Electric Field • Electric field is said to exist in the region of space around a charged object: the source charge. • Concept of test charge: • Small and positive • Does not affect charge distribution • Electric field intensity: • Existence of an electric field is a property of its source; • Presence of test charge is not necessary for the field to exist; + + + + + + + + +

  4. Some Typical Electric Fields • Near a charged, hard rubber rod • Television picture tube • Field that will create a spark in air • At electron’s orbit in hydrogen atom • 1 x 103 • 1 x 105 • 3 x 106 • 5 x 1011 Field Value (N/C)

  5. The electric field will be directly proportional to the charge setting up the field and inversely proportional to the square of the distance between that charge and where you are measuring the strength of the electric field. In other words, Electric field = (Coulomb’s constant) (Charge on object making the field) (distance away from the charge)2 E is in N/C K=9x109 Nm2 C 2 Q is in Coulombs Distance is in meters E = kq d2

  6. Example of Field Lines • Electric Field Lines

  7. Problem 1 • A positive test charge of 4.0 x 10-5 C is placed in an electric field. The force acting on the test charge is 0.60 N at 10o. What is the electric field intensity at the location of the test charge? • USE THE STEPS! • What we know: • Q = 4.0 x 10-5C • F = 0.60 N at 10o • Equation: • E = F/q

  8. Problem continued • Substitute in values: • E = 0.60 N • 4.0 x 10-5C • Solve the math: • E = 1.5 x 104 N/C at 10o • Check answer! • Reasonable answer? • Units???

  9. Problem : What is the electric field 0.2 meters away from a +4 Coulomb charge? Equation: E = kq/d2 E = (9x109 Nm2/C2)(4C) (0.2 m)2 The direction of the electric field would be away from the +4C charge since the direction is always the direction a small positive test charge would move. E=9x1011 N/C

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