A particle possessing 5.25 mu C of charge and a mass of 8.55 g is fired at a spe

Question

A particle possessing 5.25 mu C of charge and a mass of 8.55 g is fired at a speed of 357 cm/s between two horizontal charged plates of length 54.8 cm, as shown in the figure. If the electric field is constant at 3.20 times 10^3 NIC between the two plates* and directed upwards, calculate the end distance y by which the charge falls below a straight-line path. Assume a gravitational acceleration of g = 9.81 m/s^2. Number cm What field strength will allow the particle to pass between the plates along a straight path? Number N/C *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

Upward electric force, F = qE
= 5.25 x 10-6 x 3.2 x 103
= 16.8 x 10-3 N
Acceleration due to electric force = F/m
= (16.8 x 10-3) / (8.55 x 10-3)
= 1.965 m/s2.

Net acceleration, a = g - 1.965
= 7.845 m/s2.

Time taken to travel 54.8 cm is, t = distance/speed
= 0.548/3.57 = 0.154 s.

Initial downward speed, u = 0
Downward acceleration, a = 7.845 m/s2.
Time taken, t = 0.154 s
Using the eqution, y = ut + 1/2 at2.
= 0 + 0.5 x 7.845 x (0.154)2.
= 0.092 m
= 9.2 cm

b)
Downward gravitational force = Upward electric force
mg = qE
E = mg/q
= [(8.55 x 10-3) x 9.81] / (5.25 x 10-6)
= 15.98 x 103 N/C

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