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Homework Help: Parallel-plate capacitor

  1. Oct 2, 2008 #1
    A parallel-plate capacitor is formed of two 10 cm × 10 cm plates spaced 1.0 cm apart. The plates are charged to ±1.0 nC. An electron is shot through a very small hole in the positive plate.
    a. What is the slowest speed the electron can have if it is to reach the negative plate (Hint: set final speed
    at negative plate equal to zero)?

    I used C=EoA/d (8.85 x 10^-12)(100 x 10^-2)/(1 x 10^-2) = 8.85

    Then I used [tex]\Delta[/tex]V=Q/C 0 - Vi = (1 x 10^-9)/8.85

    So, Vi = -1.13 x 10^-10

    Can anyone verify this for me?
  2. jcsd
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