An electron is launched at a 45 degree angle and a speed of 5.0*10^6m/s from the positive plate of the parallel-plate capacitor shown in the figure(Figure 1) . The electron lands 4.0 cm away.
1- What is the electric field strength inside the capacitor?
Express your answer with the appropriate units.
2- What is the smallest possible spacing between the plates?
electron follows projectile path whose range is 4cm.
Range = v^2sin(2*theta) / a
0.04 = (5x10^6) sin(2x45) / a
a =6.25 x 10^14 m/s2
and a = qE /m
6.25x 10^14 = 1.6x 10-19 xE / 9.109 x 10-31
E = 3558.20 N/C
b) that will be the max. height of the projectile.
h = u^2 sin^2 (theta) / 2a
= (5x10^6) ^2 sin^2(theta) / 2x 6.25x 10^14
h = 0.01 m = 1 cm