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

To calculate the charge-to-mass ratio, the motion of the charged object in the electric field is analyzed to determine the acceleration, and then the time spent in the electric field is used to calculate the vertical deflection, which is then used to solve for the charge-to-mass ratio.

Step-by-step explanation:

We have been given the parameters of an object with mass m, charge q, initial speed v0 = 5.00 × 103 m/s, and the electric field E = 800 N/C. The condition that gravity and air resistance are negligible simplifies the problem to one dimension in the vertical plane.

The charge-to-mass ratio (q/m) can be determined by analyzing the motion of the object as it moves through the electric field. Under the influence of an electric field, a charged particle experiences a force F = qE, which results in an acceleration given by a = F/m = qE/m.

The time t the particle spends in the field can be calculated by dividing the length of the plates L by the horizontal speed v0: t = L/v0 = 0.26 m / (5.00 × 103 m/s).

We can then calculate the vertical deflection d using the equation d = 0.5 * a * t2. Substituting for a, we get d = 0.5 * (qE/m) * t2. Solving for q/m gives us:

q/m = 2d / (Et2)

Substituting the given values and solving, we can find the charge-to-mass ratio.

Answer 2

q/m = 2177.4 C/kg

Step-by-step 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