The electrostatic force acting on a particle has a magnitude of: The electrostat

Question

The electrostatic force acting on a particle has a magnitude of:

The electrostatic force exerted on a particle by the uniform electric field is directed:

During the period of time it takes the particle to travel a distance of 100 mm anti-parallel to the electric field, the work done on the particle by the electrostatic force is equal to:

The final kinetic energy of the particle is equal to:

During the period of time it takes the particle to travel a distance of 100 mm anti-parallel to the electric field, the change in the electrostatic potential energy of the system of particle plus uniform electric field, the change in the electrostatic potential energy of the system of particle plus uniform electric field is equal to:

The constant acceleration experienced by the particle has a magnitude of:

The constant acceleration experienced by the particle is directed:

How much time did it take the particle to travel the 100 mm antiparallel to the uniform electric field:

The particle's final speed is equal to:

Please answer all parts of the problem and show all work. Thanks!

A negatively charged particle of mass m= 17x10 * kg and q = -3.20x10 19 C is immersed in a uniform 2.00 x 10' NIC electric field pointing in the negative x direction, as shown in the figure. The particle is initially traveling antiparallel to the uniform electric field at a speed of 2.00x10 m/s. Assume that the gravitational force acting on the particle is negligible compared to the electrostatic force acting on it. Consider a period of time during which the particle travels a distance of 100 x10 mm antiparallel to the uniform electric field. ty uniform E (2000 NIC) 100 mm

Explanation / Answer

1) The electrostatic force acting on a particle has a magnitude of, F = q*E

= 3.2*10^-19*2000

= 6.4*10^-16 N

2) Direction of F: towards +x axis

3) Workdone by elctroc static force, W = F*d

= 6.4*10^-16*100*10^-3

= 6.4*10^-17 J

4) Apply, W = KEf - KEi

==> KEf = KEi + W

= 0.5*m*vi^2 + W

= 0.5*1.17*10^-24*(2*10^4)^2 + 6.4*10^-17

= 2.98*10^-16 J

5) change in potentail energy = -workdone

= -2.98*10^-17 J

6) acceletaion, a = F/m

= 6.4*10^-16/(1.17*10^-24)

= 5.47*10^8 m/s^2

7) direction of a : towards +x axis

8) Apply, d = vo*t + 0.5*a*t^2

0.1 = 2*10^4*t + 0.5*5.47*10^8*t^2

==> 2.735*10^8*t^2 + 2*10^4*t - 0.1 = 0

on solving the above equation we get'

t = 4.7*10^-6 s

9)
vf = vi + a*t

= 2*10^4 + 5.47*10^8*.7*10^-6

= 2.038*10^4 m/s

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