0036

Question : A particle of charge 2µC and mass 1.6g is moving with a velocity 4i m/s. At t=0 the particle enters in a region having an electric field E (in N/C)=80i+60j. Find the velocity of the particle at t=5s.

Doubt by Safir

Solution : 

q=2µC
  = 2×10^-6C
m=1.6 g
m=1.6×10^-3 Kg

u_x = 4 m/s
u_y= 0 m/s

t=5s

E (in N/C)=80i+60j
E_x=80i N/C
E_y=60 j N/C

We know, 
F=qE
ma=qE
a=qE/m
a_x=qE_x/m
a_x=80q/m

a_y=qE_y/m
a_y=60q/m

Now 
v_x=u_x+axt
v_x=4+(80q/m)5
v_x=4+(400q/m)
v_x=4+[(400×2×10^-6)/1.6×10^-3]
v_x=4+[800/1.6]×10^-3
v_x=4+[500×10^-3]
v_x=4+0.5
v_x=4.5 m/s

v_y=u_y+a_yt
v_y=0+(60q/m)5
v_y=300q/m
v_y=[(300×2×10^-6)/1.6×10^-3]
v_y=[600/1.6]×10^-3
v_y=[375×10^-3]
v_y=0.375
v_y=0.375 m/s

Hence, final velocity of the particle at t=5s will be
=v_xi+v_yj
=(4.5i+0.375j) m/s