onstants Part A A small object with mass m, charge q, and initial speed vo 5.00x

Question

onstants Part A A small object with mass m, charge q, and initial speed vo 5.00x103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1). The electric field between the plates is directed downward and has magnitude 600 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= 1.35 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. g/m = 9.13-103 C/kg revious Answer All attempts used; correct answer withheld by instructor

Explanation / Answer

consider the motion along the horizontal direction

vo = initial velocity = 5000 m/s

X = total distance travelled before hitting the collecting plate = 56 + 26 = 82 cm = 0.82 m

t = total time of travel = X/vo = 0.82/5000 = 0.000164 sec

tp = time of travel between the plates = 0.26/5000 = 0.000052 sec

consider the vertical motion inside the plates

a = acceleration of the charge particle = qE/m

vo = initial velocity = 0 m/s

yp = deflection between the plates = ?

tp = time of travel between the plates = 0.000052 sec

using the equation

yp = vo tp + (0.5) a tp2

yp = (0) (0.000052) + (0.5) (qE/m ) (0.000052 )2

yo = deflection outside = d - yp

velocity gained by the particle is given as

v = 0 + at = (qE/m ) (0.000052 )

to = time of travel outwise the plates = 0.56/5000

also

to = yo /v

(0.56/5000) v = (d - yp)

(0.56/5000) (qE/m ) (0.000052 ) = (0.0135 - (0.5) (qE/m ) (0.000052 )2 )

qE/m = 1.88 x 106

q(660)/m = 1.88 x 106

q/m = 2848.5

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