A particle possessing 3.25 muC of charge and mass 3.55 g is fired at a speed of

Question

A particle possessing 3.25 muC of charge and mass 3.55 g is fired at a speed of 459 cm/s through two charged plates of length 44.7 cm, as shown in the figure. If the electric field is constant at 2140 N/C between the two plates* and directed upwards, calculate the distance y in which the charge falls below its intended path. Number cm v = 459 cm/s Which field strength will allow the particle to pass between the plates along a straight path. * In actuality, the electric field will vary considerably near the edges of the plates. We will ignore this situation and consider the electric field constant everywhere between the plates.

Explanation / Answer

a)Net downward acceleration=g-qE/m=7.841m/s2

time to cross plate=44.7/459=0.0974s

displacement=0.5at2=3.72cm

b)Balancing forces,

mg=qE

E=mg/q=10716N/C

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