o 2/2/2018 11:59 PM , 663/1002/2/2018 11:33 PM Gradebo Print CalulatorPeriodic T
ID: 1577311 • Letter: O
Question
o 2/2/2018 11:59 PM , 663/1002/2/2018 11:33 PM Gradebo Print CalulatorPeriodic Table Solution Map Sapling Learning s between two horizontal charged plates of length 34.6 cm, as shown in the between the two plates and directed upwards, calculate the end distance y by which the charge fails below a e. If the electric field is constant at 1770 N/C straight-line path. Assume a gravitational acceleration of g 9.81 m/s? Numbor L-34.6 2.23 cm v 459 cm/s What field strength will allow the particle to pass between the plafes along a straight path? Number 8860 N/C In actuality, the electric field wil vary considerably near the odges of the plates. We will ignore this situation and consider the electric field constant everywhere between the plates 0 View Preveus View Solton, 0 View Next Exa hp ,' '21/1/1/1/vv1/1/1/10000000000 ' 1/1/ , 0000Explanation / Answer
a)
Net force in y-direction,
Fnet = Fg - Fe
m*ay = m*g - q*E
ay = g - q*E/m
= 9.81 - 7.25*10^-6*1770/(6.55*10^-3)
= 7.85 m/s^2
time taken for the chrged particle to cross the plates,
t = L/vo
= 0.346/4.59
= 0.07538 s
deflection y direction during this time,
y = (1/2)*ay*t^2
= (1/2)*7.85*0.07538^2
= 0.0223 m
= 2.23 cm
b) use, Fe = Fg
q*E = m*g
E = m*g/q
= 6.55*10^-3*9.81/(7.25*10^-6)
= 8863 N/c
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