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?