Protons are projected with an initial speed of Vi kmsec int

Protons are projected with an initial speed of V_i = km/sec into a uniform electric field E as shown in the figure. The protons are to hit a target at a horizontal distance of 1.27 mm from the point where the protons cross the plane. Find the projection angles theta that will result in a target hit and find the time of flight (or time the proton is above the plane) for each trajectory.

Solution

E = -720 j N/C

Force on the proton downwards = 1.6 *10^-19 *720 = 1.152*10^-16 N

Acceleration due to E downwards = 1.152 *10^-16 / (1.67*10^-27) = 0.69 * 10^11 m/s^2

Range is u^2 sin2(theta) / a

So 1.27 * 10^-3 = (9.67*10^3)^2 sin2(theta) / (0.69 *10^11)

sin2(theta) = 0.9371

theta = 69.57/2 = 34.79 degrees

Time of flight = 2 * 9.67 * 10^3 * sin (34.79) / 0.69 * 10^11) = 16 * 10^-8 s