Question

A small object with mass m, charge q, and initial speed v0 = 5.00 * 103 m>s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Fig. P21.78). The electric field between the plates is directed downward and has magnitude E = 800 N>C. Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.25 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object’s charge-to-mass ratio, q>m

Answer 1

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