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 21 volts and the distance between the plates is 12 milli-meter (no dielectric). A uniform magnetic field of 88 milli-Tesla is in the -z direction. A point charge of -22 micro-coulombs with a mass of 5 grams is passing through at 23 percent of the speed not to be deflected. It is initially moving horizontally in the +x direction. If the length of the plates is 5 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

Force on -q due to Electric field of capacitor in +Y-direction = qE = q(V/d) = 22 X 10^(-6) C X [21V/(12X10^(-3) m]

=3.85 X 10^(-2) N

If the charge is not be deflected from it's path along +X-axis , a force of magnitude 3.85 X 10^(-2) N is required to act on it in (-) Y-direction.

In this case the above force is produced due to Force on q due to B-field.

Since charge has velocity along +X-axis and B is into screen, the force due to B is in vertically downward direction and has a magnitude

F = q V X B = qvB in (-) Y-direction

= 22 X 10^(-6) C X v X 88 X 10^(-3)

If E force and B force are balanced

22 X 10^(-6) C X v X 88 X 10^(-3) T = 3.85 X 10^(-2) N

v = 3.85 X 10^(-2) N/[ 22 X 10^(-6) C X 88 X 10^(-3) T] = 1.989 X 10 ^(+4) m/s

If the charge has 23 % of this speed , speed v' = 23 X 1.989 X 10^4/100 =45.747 X 10^2 m/s

At this speed net force on -q is towards +Y-direction

and it's magnitude = (1-0.23 ) X 3.85 X 10^(-2) N = 0.77 X 3.85 X 10^(-2) N

= 2.96 X 10^(-2) N

Acceleration of chaarge = 2.96 X 10^(-2) N/(5X 10^(-3) = 5.93 m/s^2

Horizonal Distance travelled by charge to clear the plates = 5 mrs

Horizontal velocity = 4.57 X 10^3 m/s

time for clearing the plates along horizontal direction = 5/ [ 4.57 X 10^3 ] = 1.09 X 10^(-3) secs

In this time vertical distance travelled = 0.5 X 5.39 m/s^2 X 1.09 X 10^(-3) secs

= 2.95 X 10^(-3 ) mrs = 2950 micro meters

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