1. The problem statement, all variables and given/known data An electron is launched at a 45∘ angle and a speed of 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor shown in the figure (Figure 1) . The electron lands 4.0 cm away. a) What is the electric field strength inside the capacitor? b) What is the smallest possible spacing between the plates? 2. Relevant equations F = qE Equations used for Projectile Motion 3. The attempt at a solution For Part A Okay so first I thought that I may need the acceleration in the y-direction to use F = qE, so I separated the velocity into separate components. Vx = 5.0×10^6 m/s * cos(45) Vy = 5.0×10^6 m/s * sin(45) Then I found the time. 0.04m = (5.0×10^6 m/s * cos(45))*t t = 1.131*10^-8 seconds Then I proceeded to finding the acceleration with the time I found by using the highest point reached (0 m/s) 0 m/s = 5.0×10^6 m/s * sin(45) + a (1.131*10^-8 sec) a = -2.69 * 10^4 m/s^2 After that I found the force on the electron using its mass, 9.10938356 × 10^-31 kilograms F = (9.10938356 × 10^-31 kg)*(-2.69 * 10^4 m/s^2) F= -2.459*10^-16 N Then used that to find E in F= qE using the charge of the electron 1.6*10^-19 C 2.459*10^-16 N = (1.6*10^-19 C)*(E) E = -1536.57 N/C But it's wrong. Can someone tell me what I did wrong? For Part B. I tried calculating the amount of distance the electron traveled in the vertical direction. 0^2 = (5.0×10^6 m/s * sin(45))^2+2(-2.69 * 10^4 m/s^2)*y y= 0.023 m But it's wrong. I think I might have misunderstood the question. How do I find the answer? Any help is much appreciated!