Answer:
q/m = 2177.4 C/kg
Explanation:
We are given;
Initial speed v_o = 5 × 10³ m/s = 5000 m/s
Now, time of travel in electric field is given by;
t_1 = D_1/v_o
Also, deflection down is given by;
d_1 = ½at²
Now,we know that in electric field;
F = ma = qE
Thus, a = qE/m
So;
d_1 = ½ × (qE/m) × (D_1/v_o)²
Velocity gained is;
V_y = (a × t_1) = (qE/m) × (D_1/v_o)
Now, time of flight out of field is given by;
t_2 = D_2/v_o
The deflection due to this is;
d_2 = V_y × t_2
Thus, d_2 = (qE/m) × (D_1/v_o) × (D_2/v_o)
d_2 = (qE/m) × (D_1•D_2/(v_o)²)
Total deflection down is;
d = d_1 + d_2
d = [½ × (qE/m) × (D_1/v_o)²] + [(qE/m) × (D_1•D_2/(v_o)²)]
d = (qE/m•v_o²)[½(D_1)² + D_1•D_2]
Making q/m the subject, we have;
q/m = (d•v_o²)/[E((D_1²/2) + (D_1•D_2))]
We have;
E = 800 N/C
d = 1.25 cm = 0.0125 m
D_1 = 26.0 cm = 0.26 m
D_2 = 56 cm = 0.56 m
Thus;
q/m = (0.0125 × 5000²)/[800((0.26²/2) + (0.26 × 0.56))]
q/m = 312500/143.52
q/m = 2177.4 C/kg
