A particle possessing 8.25 mu C of charge and a mass of 7.55 g is fired at a spe

Question

A particle possessing 8.25 mu C of charge and a mass of 7.55 g is fired at a speed of 459 cm/s between two horizontal charged plates of length 34.6 cm, as shown in the figure. If the electric field is constant at 1.80 times 10^3 N/C 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. What 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)

Time taken to cross the plate

t=L/v=34.6/459 =0.07538 s

from newtons 2nd law ,net force acting on particle is

Fnet=mg-FE

ma=mg-qE

a=g-(qE/m) =9.81-(8.25*10-6)*1800/(7.55*10-3)

a=7.843 m/s2

from

Y=Vot+(1/2)at2=0+(1/2)*7.843*0.075382

Y=2.23 cm

b)

E=mg/q=(7.55*10-3)*9.81/(8.25*10-6)

E=8977.6 N/C

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