In the following velocity selector, the positive plate of the capacitor is above

Question

In the following velocity selector, the positive plate of the capacitor is above the negative as shown. The voltage of the battery is 25 volts and the distance between the plates is 16 milli-meter (no dielectric). A uniform magnetic field of 51 milli-Tesla is in the -z direction. A point charge of -23 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. -q Uniform magnetic field in the z direction (into the plane) You Answered 5.3910 Correct Answer 452 margin of error +/-10%

Explanation / Answer

Fe = q E

E = V/d = 25 / (16 x 10^-3) = 1562.5 V/m downward  

Fe = (-23 x 10^-6) (- 1562.5) = 0.036 N upward  

Fb = 0.16 (0.036) = 0.00576 N downward  

Fnet = Fe - Fb = 0.0302 N upward

not to be deflected speed = E/B = (1562.5)/(51 x 10^-3)

= 30637 m/s  

v = (0.16)(30637) = 4900 m/s

time taken to cross, t = (6)/4900 = 1.224 x 10^-3 sec  

a = F / m = (0.0302)/(5 x 10^-3) = 6.04 m/s^2

d = a t^2 /2 = (6.04)(1.224 x 10^-3)^2 / 2

d = 4.52 x 10^-6 m

Ans: 4.52

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