A positive point charge q+2pC of mass m-200g is released from rest in a uniform electric field E-2x10 N/C directed along the s-axis as shown in figure, the distances between tow plate is 20cm, answer Q, Q20, and Q21. Q15) The acceleration of charge in m/s² is: a) 10 b) 20 c) 30 d) 40 Q16) The speed of the point charge when it reached to the negative plate in m/s is: a) 1.8 b) 2 c) 2.3 d) 2.8 Q17) The kinetic energy of charge when it reached to the negative plate in joul is: a) 0.4 b) 0.6 c) 0.8 d) 1.6 13 00000

Charge on particle = 2×10^-6 C.

Mass = 200 gm = 200 × 10^-3 kg = 0.2 Kg 

Electric field = 2 × 10^6 N/C. 

Force on charge = E × Q = (2×10^-6) × ( 2 × 10^6) 

Force = 4N.( In the direction of electric field). 

Acceleration= (force /mass ) =( 4/0.2)= 20m/s^2. 

Distance between plate = 20cm = 0.20 m 

Use the relation:- 

V^2 = U^2 + 2× a × s 

U = initial velocity = 0m/s 

a = acceleration = 20m/s^2 

S = distance = 0.20m 

Putting the value :- 

V^2 = (2× 20 × 0.20 ) = 8 

Hence , V = 8^0.5 = 2.828 m/s^2