Question

A point charge of -3.5 micro-coulombs with a mass of 0.42 kg is placed in a uniform electric field directed in the +x direction. Initially, the charge is moving in +x direction at 9.5 m/s and after it has traveled a displacement of 2.2 meters in the -x direction, it is moving at 24.7 m/s in the -x direction. If a piece of paper with an area of 0.5 m2 is placed in the field and the angle between the normal and the field is 51 degrees, what is the electric flux through the paper in terms of 10^5 N m2/C? Ignore the weight force.

Answer 1

In the electric field

electric field = E

electric force F = E*q

q = charge = 3.5 micro coulomb

acceleration ax = F/m

m = mass of charge = 0.42 kg

acceleration ax = E*3.5*10^-6/0.42 = E*8.33*10^-6 m/s^2

initial velocity vox = 9.5 m/s

final velocity vx = 24.7 m/s

displacement x = 2.2 m

from equation of motion

vx^2 - vox^2 = 2*ax*x

24.7^2 - 9.5^2 = 2*E*8.33*10^-6*2.2

E = 142*10^5 N/C

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Electric flux = E*A*costheta

ELectric flux = 142*10^5*0.5*cos51 = 44.7*10^5 Nm^2/C <<<<-------ANSWER