A particle possessing 4.25 uC of charge and a mass of 6.55 g is fired at a speed

Question

A particle possessing 4.25 uC of charge and a mass of 6.55 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 3020 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/s2 Number K >l L=54.8 cm cm v = 357 cm/s What field strength will allow the particle to pass between the plates along a straight path? Number NIC

Explanation / Answer

v = 357 cm/s = 3.57 m/s

Force on the charged particle, F = q*E

m*a = q*E

a = q*E/m

= 4.25*10^-6*3020/(6.55*10^-3)

= 1.96 m/s^2

net acceleration, ay = g - a

= 9.8 - 1.96

= 7.84 m/s^2

time taken to travel, t = 54.8/357

= 0.1535 s

so, y = voy*t + (1/2)*ay*t^2

= 0 + (1/2)*7.84*0.1535^2

= 0.0923 m

= 9.23 cm <<<<<<<-------------------------Answer

use, Fe = Fg

q*E = m*g

E = m*g/q

= 6.55*10^-3*9.8/(4.25*10^-6)

= 15103 N/c <<<<<<<-------------------------Answer

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