1. The problem statement, all variables and given/known data The electric field strength is 2.50×104 N/C inside a parallel-plate capacitor with a 1.20 mm spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate? 2. Relevant equations F = eE = ma vf2 = vi2 + 2ad 3. The attempt at a solution First, I used F = eE to figure out the force acting on the electron and then divided the force by the mass to get the electron's acceleration in the electric field: a = eE/m = (1.60×10-19 C)(2.50×104 N/C) / (1.67×10-27 kg) = 2.3952×1012 m/s2 Then I used a kinematic equation to figure out the final velocity: vf2 = vi2 + 2ad (vi = 0 since the electron starts at rest) vf = √(2ad) = √(2(2.3952×1012 m/s2)(1.2×10-3 m)) = 7.58×104 m/s But apparently this is not the answer. Can anyone please tell me what I'm doing wrong? I've checked over my work several times. Any help would be much appreciated!