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 19 volts and the distance between the plates is 17 mill-meter (no diclectric, A uniform magnetic feld of 90 mill-Tesla is in the z direction. A paint charge of -18 micro- coulombs with a mass of 9 grams is passing through at 25 percent of the speed not the length of the plates is 3 meters, how far will the charge be deflected (vertically) in micro-meters (1 x 10 m) by the time it eits the velocity selector? ignore the weight to be deflected. It is initially moving horizontally in the ex direction If -q Uniform magnetic field in the -z direction (into the plane) 874.8000 t Answer 0.78 margin of error 10%

Explanation / Answer

Given,

V = 19 V ; d = 17 mm ; B = 90 mT ;

q = -18 uC ; m = 9 g ; v' = 25 % v

L = 3 m ;

Let the deflection be s

We know that,

E = V/d = 19/(17 x 10^-3) = 1.12 x 10^3 N/C

Fe = Fb

q E = q v B

v = E/B = 1.12 x 10^3/(90 x 10^-3) = 1.24 x 10^4 m/s

v' = 0.25 x 1.24 x 10^4 = 3.1 x 10^3 m/s

t = d/v = 3/3.1 x 10^3 = 0.96 x 10^-3 s

Now, a = Fnet/m

Fe = q E = 18 x 10^-6 x 1.12 x 10^3 = 0.0202

Fm = 0.25 x 0.0202 = 0.0051

Fnet = 0.0202 - 0.0051 = 0.0151

a = 0.0151/9 x 10^-3 = 1.68 m/s^2

s = ut + 1/2 at^2

s = 0 + 0.5 x 1.68 x (0.96 x 10^-3)^2 = 0.78 x 10^-6 m

Hence, s = 0.78 micro meter

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