Problem 21.78 14th Edition. A small object with mass m , charge q, and initial s
ID: 2003217 • Letter: P
Question
Problem 21.78 14th Edition. A small object with mass m , charge q, and initial speed vsub 0 = 6.00 x 10^3 m/s is projectected into a uniform electric field between two parallel metal plates of length 26.0 cm. The electric field between th 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 region, the object is deflected downward a vertical distance of d = 1.35 cm from its original direction of motion and reached 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 in C/kg. How do I set up this problem. I need the answer in 3 hours. Thank you .
Explanation / Answer
We assume the object is projected into the plates in horizontal direction i.e. parallel to the plates as nothing is mentioned.
Force on the particle due to electric field F = Eq = 800q N
The force acts downward as the field is directed downward
Acceleration a = F/m = 800q/m
Horizontal velocity v0 = 6.0e+3 m/s
X = vo t = 0.26, where t is time flight across the plates
t = 0.26/6e+3 = 4.33e-5 s
vertical velocity of the object at the end of the plates v = at = (800q/m)*4.33e-5 m/s
vertical displacement across the plates = at2/2 = (800q/2m)(4.33e-5)2
time of flight after the plates t2= 0.56/6e+3 = 9.33e-5 s
vertical distance travelled after the plates vt2 = (800q/m)*4.33e-5 *(9.33e-5)
total vertical distance = 1.35cm
total vertical distance = 1.35cm
(800q/2m)(4.33e-5)2 = 1.35e-2
q/m = 1.8 e+4
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there is a little consfusion, if the vertical disatance 1.35 cm is at the end of the plates region we do not need the rest of the information for q/m
on the other hand if the distance 1.35 cm is till it reached the collecting plates
vertical distance travelled after the plates vt2 = (800q/m)*4.33e-5 *(9.33e-5)
total vertical distance = 1.35cm
(800q/2m)(4.33e-5)2 + (800q/m) *4.33e-5 *(9.33e-5) = 1.5e-2
q/m *3.98e-6 = 1.35-2
q/m = 3.39e+3
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