I need help with all the questions below including 21 pleas

I need help with all the questions below including # 21, please.

Also...

21. The work done on the charged particle by the electric field equals? ???

A negatively charged particle of mass m=1.17x10-24 kg and q =-160 x 10-19 C\'s uniform 6.00x10 N/C 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 2.00x 10 m/s (at the instant when the particle is passing through equipotential surface labeled\"A\") acting on it anti-parallel to the uniform electric field, between the equipotential surfaces labeled \"A\" and \"B\" in the Figure. in a Assume that the gravitational force acting on the particle is negligible compared to the electrostatic force In the following, consider the period of time during which the particle travels a distance of 10.0 mm 16 16. The change in the particle\'s kinetic energy is equal to a. 9.60 x 10J. b. 960x10*J. c. -60.0J. d. 9.60x 10- J . c..60.01, d. 960x10-11. e. 6001 17. As it passes through equipotential surface \"B\", the particle\'s speed is equal to a. 452 x 103 m/s. b. 336x 10 m/s. c. 4.05x10 m/s. d. zero. e. 2.04 x 10 m/s. 18. The electric force acting on the charged particle has magnitude a. 9.60x 10-1 . b. 60.0 N. C c. 9.60x 10v. d. 9.60x 10\" v e. 96.0 kN 19. The electric force acting on the charged particle is directed a. in the +x direction. b. in the -x direction. c. in the +y direction. d. in the -y direction. 20. Find the time it takes the charged particle to travel the 10.0 mm from surface \"A\" to surface \"B\". a. 2.44 ps. b. 7.94 msc. 5.51. d. 3.07 us. e. 7.94 us. 4A

Solution

16)

Change in KE = E*d*q

where

E = 6*10^3 N/C

d = 10 mm = 0.01 m

q = 1.6*10^-19 C

So, change in KE = 6*10^3*0.01*1.6*10^-19

= 9.6*10^-18 J <---- option d

17)

Now, change in KE

= 0.5*m*(v^2 - u^2) = 9.6*10^-18

So, 0.5*1.17*10^-24*(v^2 - (2*10^3)^2) = 9.6*10^-18

So, v = 4.52*10^3 m/s <----- option a

18)

Force, F = E*q

= 6*10^3*1.6*10^-19

= 9.6*10^-16 N <---- option c

19)

direction = in the +x direction <----- option a

20)

F = m*a

So, a = F/m

= 9.6*10^-16/(1.17*10^-24)

= 8.21*10^8 m/s2

So, using the equation of motion,

s = ut + 0.5*at^2

u = 2*10^3 m/s

So, 0.01 = 2*10^3*t + 0.5*(8.21*10^8)*t^2

So, t = 3.07*10^-6 s = 3.07 us <---- option d

21)

Work done = change in KE = 9.6*10^-18 J