Electric and Magnetic Fields

Force acting on the charge is

$\vec F=q(\vec E+\vec v\times \vec B) =q\left (\hat j(E-v_xB)+\hat iv_yB\right )$ .

So

$\dfrac{dv_x}{dt}=\dfrac{qB}{m}v_y, \dfrac{dv_y}{dt}=\dfrac{qE}{m}-\dfrac{qB}{m}v_x\implies \dfrac{d^2v_y}{dt^2}=-(\frac{qB}{m})^2v_y$ .

Solution of the last equation is

$v_y=A\sin (\frac{qBt}{m})$ , where A, the amplitude of $v_y$ , is the integration constant. The other constant is zero since initial velocity is zero in both the $x$ and the $y$ directions. Since there is no acceleration in the $z$ direction and the initial velocity in this direction is zero, there is no motion along this direction. So,

$v_x=\dfrac{m}{qB}(\frac{qE}{m}-\frac{dv_y}{dt})=\dfrac{E}{B}-A\cos (\frac{qBt}{m})$ .

Since at $t=0, v_x=0$ ,

Therefore $A=\dfrac{E}{B}\implies v_y=\dfrac{E}{B}\sin (\frac{qBt}{m})\implies y=\dfrac{mE}{qB^2}\left (1-\cos (\frac{qBt}{m})\right )$

$\implies y_{max}=\dfrac{2mE}{qB^2}$ .

Substituting values, we get $y_{max}=\boxed {0.64}$

Just for fun, here's my general computational solution. If we wanted to, we could make another version of this problem with all of the field components populated, and all of the initial velocities populated.

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import math

m = 2.0
q = 1.0
E = 4.0
B = 5.0

dt = 10.0**(-5.0)

##########################

Ex = 0.0
Ey = E
Ez = 0.0

Bx = 0.0
By = 0.0
Bz = B

##########################

t = 0.0
count = 0

x = 0.0
y = 0.0
z = 0.0

xd = 0.0
yd = 0.0
zd = 0.0

xdd = 0.0
ydd = 0.0
zdd = 0.0

ymax = 0.0

##########################

while t <= 10.0:

    x = x + xd*dt
    y = y + yd*dt
    z = z + zd*dt

    xd = xd + xdd*dt
    yd = yd + ydd*dt
    zd = zd + zdd*dt

    FEx = q*Ex
    FEy = q*Ey
    FEz = q*Ez

    crossx = yd*Bz - zd*By
    crossy = -(xd*Bz - zd*Bx)
    crossz = xd*By - yd*Bx

    FBx = q*crossx
    FBy = q*crossy
    FBz = q*crossz

    Fx = FEx + FBx
    Fy = FEy + FBy
    Fz = FEz + FBz

    xdd = Fx/m
    ydd = Fy/m
    zdd = Fz/m

    t = t + dt
    count = count + 1

    if y > ymax:
        ymax = y

    #if count % 1000 == 0:
        #print t,x,y,z

##########################

print dt
print ymax

#>>> 
#0.0001
#0.640880858483
#>>> ================================ RESTART ================================
#>>> 
#1e-05
#0.640087976691
#>>> 

I will be happy if anyone will make a nice follow up to this problem.