A particle with positive charged of q=20 nC and mass m=10^-9 kg enters the space

Question

A particle with positive charged of q=20 nC and mass m=10^-9 kg enters the space between two parallel plates at (x,y)=(0,0) going at velocity v0 = 100 m/s in the +y direction. A voltage Vp of 20 V is applied across the plates to create surface charge densities on the plates that are equal and opposite, as shown. The plates are 0.2m apart and located at x= +/- d/2. The plates extend to 20 meters in the y direction. At what y value does the electron hit one of the plates? (Next question asks the x value)

Explanation / Answer

charge of the particle = +20*10-9 C

mass of the charge particle = 10-9 kg

distance between the parallel plate = d = 0.2 m

connected by 20 V voltage source. therefore potential difference between two plates = 20 V

hence electric field generated due to this voltage between plates = V/d = 20/0.2 = 100 V/m

forces acting on the charge particles:

1). gravitational force in -ve y-direction

2). electrical force acting on the charge particle in +ve x-direction

particle is initially moving with 100 m/s of velocity starting from the mid of the parallel plate.

g= 9.81 m/s2

+ve charge attracts to the -ve plate. so charge particle start tending to hit towards -ve charged plate.

FE = q*E = +20*10-9 *100 = 2000*10-9 N

acceleration of the charge particle due to electric field = F/m = 2000*10-9/10-9 = 2000 m/s2

both accelerations are perpendicular. therefore, resultant acceleration = (a12+g2) = let a m/s2

from the newton's 3rd law of the motion,

v2 - u2 = 2ah

v= 0 m/s ; u = 100 m/s

height = h m ; acceleration = - a m/s2

therefore,

02 - 1002 = - 2*(20002+9.812)*h

after solving this equation, we get;

h = 2.5 m

hence, at y = 2.5 m the electron will hit one of the plates.

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