a) Calculate the magnitude of the electric field, which is uniform, between the

Question

a) Calculate the magnitude of the electric field, which is uniform, between the two plates, E

b) and the force acting on the charged ink drop by the field, F, and its acceleration in the y direction, ay.

c) the time which takes the drop to move out of the plates, t

d) the y component of its velocity when the drop leaves the plates, vy

(0 = 40 4- degrees) (alpha) A charged particle is projected in a uniform electric field with initial velocity V0 = 5 Times 103m / s at theta = 40 degree to the x-axis. Its charge is q = - 1.6 Times 10-11 C. and mass is m = 3.0 Times 10 -14 kg. The two parallel plates are separated by d = 2.0 cm. The electric potential difference between the plates is 3,600 volts. The length of the plates is 10.0 cm. Calculate the magnitude of the electric field, which is uniform, between the two plates, E, and the force acting on the charged ink drop by the field. F, its acceleration in y-direction, ay, the time which lakes the drop to move out of the plates, t, the y-component of its velocity when the drop leaves the plates, vy its velocity and the angle alpha between the velocity of the particle and the x-axis when it leaves the plates.

Explanation / Answer

(a) magnitude of electric field = V/d = 3600/0.02 = 180,000 V/m

(b) F = E*q = 180000 * -1.6*10^-11 = -2.88*10^-6 N
a = F/m = -96*10^6 m/s^2

(c) velocity component in x-direction is constant = 5000*cos40 = 3830.22 m/s
Time taken = d/t
= 0.1/3830
= 2.6*10^-5 s

(d) v in y-direction = v sin40 + a*t
5000sin40 - (96*10^6 * 2.6*10^-5)
= 707.55 m/s

(e) tan(alpha) = vy/vx = 707.55/3830
alpha = 10.46 degrees

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