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For more information, please see full course syllabus of AP Physics B
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  • Capacitance: If we have 2 conductors and we place charge +Q on one and –Q on the other, a potential difference V develops, the conductor with positive charge being at a higher potential. The capacitance of this system, consisting of the two conductors, is C = Q / V.
  • For a parallel plate capacitor, C = (epsilon_0) A / d, where A is the area of one of the plates (the two plates have the same area A) and d is the separation between the plates.
  • If a dielectric (an insulator) with dielectric constant K fills the space between the plates of a parallel plate capacitor, then the capacitance becomes KC, where C is the capacitance in the absence of the dielectric.
  • The energy stored in a capacitor is Q^2 / 2C, where Q is the charge on one plate. It is also given by (1/2) C V^2, where V is the potential difference across the plates.


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
  • Capacitance 0:09
    • Consider Two Conductor s
    • Electric Field Passing from Positive to Negative
    • Potential Difference
    • Defining Capacitance
  • Parallel Plate Capacitance 8:30
    • Two Metallic Plates of Area a and Distance d
    • Potential Difference between Plates
  • Capacitance with a Dielectric 22:14
    • Applying Electric Field to a Capacitor
    • Dielectric
  • Example 34:56
    • Empty Capacitor
    • Connecting Capacitor to a Battery
    • Inserting Dielectric Between Plates
  • Energy of a Charged Capacitor 43:01
    • Work Done in Moving a Charge, Difference in Potential
  • Example 54:10
    • Parallel Plate Capacitor
    • Connect and Disconnect the Battery
    • Calculating Q=cv
    • Withdraw Mica Sheet
    • Word Done in Withdrawing the Mica
  • Extra Example 1: Parallel Plate Capacitor
  • Extra Example 2: Mica Dielectric