Question 4 22 pts In the following velocity selector, the positive plate of the

Question

Question 4 22 pts In the following velocity selector, the positive plate of the capacitor is above the negative as shown. The voltage of the battery is 29 volts and the distance betweer the plates is 15 milli-meter (no dielectric). A uniform magnetic field of 96 milli-Tesla is in the -z direction. A point charge of -20 micro-coulombs with a mass of 5 grams is passing through at 16 percent of the speed not to be deflected. It is initially moving horizontally in the +x direction. If the length of the plates is 6 meters, how far will the charge be deflected (vertically) in micro-meters (1 x 106 m) by the time it exits the velocity selector? Ignore the weight. 9 Uniform magnetic field in the -z direction (into the plane)

Explanation / Answer

Question 4 :

V = potential difference between the plates = 29 Volts

d = distance between the plates = 15 mm = 0.015 m

E = electric field between the plates = V/d = 29/0.015 = 1.93 x 103 V/m

B = magnetic field = 96 mT = 0.096 T

undeflected speed is given as

v = E/B = (1.93 x 103)/0.096 = 2.01 x 104 m/s

v' = speed of travel of charge along horizontal direction = 0.16 v = 0.16 x 2.01 x 104 = 3216 m/s

L = length of the plate = 6 m

t = time taken to travel the length of plate = L/v' = 6/3216 = 1.87 x 10-3 sec

a = acceleration due to electric field = qE/m = (20 x 10-6) (1.93 x 103)/(5 x 10-3) = 7.72 m/s2

deflection in vertical direction is given as

Y = (0.5) a t2

Y = (0.5) (7.72) (1.87 x 10-3)2

Y = 1.35 x 10-5 m = 13.5 x 10-6 m

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