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For more information, please see full course syllabus of AP Physics C/Electricity and Magnetism
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Lecture Comments (1)

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Post by Prachum Chanman on February 19, 2013

A parallel plate capacitor has 3.1 J of energy stored in it. The seperation between the plates is 7.70 mm. Calculate the amount of energy stored in the plates if the seperation between the plates was decreased to 4.41 mm while the potential difference across the plates remains constant.

Electric Potential, Part 1

  • The potential difference between two points A and B, written as V_b – V_a, is equal to the work done in moving a unit charge from A to B.
  • If a uniform E-field exists in a region of space, the V_b – V_a = - E.d, the dot product of the vector E and the vector d from A to B.
  • In a parallel plate capacitor, the potential difference between the plates is equal to Ed, where E is the electric field between the plates and d is the separation between the plates. The positive plate is at a higher potential than the negative plate.

Electric Potential, Part 1

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Potential Difference Between Two Points 0:16
    • Electric Field in Space By Stationary Charges
    • Point Charge Moves From A to B
    • Electric Field Exerts a Force
    • Electric Potential Energy
    • Work Done By External Agent
    • Change in Potential Energy is Equal to Amount of Work Done
  • Potential Difference in Uniform Electric Field 27:59
    • Constant Electric Field
    • Equipotential
  • Parallel Plates 40:52
    • Electric Field is Perpendicular to Plate
    • Charge Released at A from Rest
  • Motion of Charged Particle in a Uniform Electric Field 51:55
  • Example 1: Work by Moving Electrons
  • Example 2: Block and Spring
  • Example 3: Particle on String