How do you solve?

For a central field given by

#vec E = k_0 vec r# we have

#vec F = q vec E# (Force central field)

#V(r) = -int F(r)dr = -q k_0 int r dr = -1/2 q k_0 r^2# and also

#U = 1/2m v^2+V(r) = 1/2m v^2-1/2qk_0 r^2 = C^(te)# so

#1/2 m v_1^2-1/2qk_0 r_1^2 = -1/2qk_0 r_2^2#

where #r_1 = l# and #r_2 = l/sqrt2# and then

#m v_1^2-qk_0 l^2=-qk_0 l^2/2# then

#v_1=sqrt(qk_0/(2m)) l=sqrt((lqE_0)/(2m))# but

#(q E_0l)/m=8# then

#v_1= 2# (Minimum velocity in consistent units)

It is given that the ring is projected along the rod from point #(0,l)#.

Electric force on charge #+Q# is given as

#vecF=QvecE#

We observe that at the initial point #(0,l)#, component of electric force along the direction of projection opposes motion. As the ring moves towards point #(l,0)#, this component decreases, becomes zero at #(l/2,l/2)#. Thereafter this component aids motion along the rod towards #(l,0)#. The orthogonal component of electric force on does not affect the motion of the ring.

From the foregoing we infer that initial kinetic energy given to the ring must be able to just pass the point #(l/2,l/2)#.

Work done by the charge to move a distance #dvecs# along the direction of projection is given by

#dw=vec(F)*dvecs#

Inserting given values and noting the negative slope of rod we get

#dw=(QE_0)/l(xhat i+yhatj)cdot (dxhat i-dyhatj)=(QE_0)/l(xdx-ydy)#

Total work done to reach point #(l/2,l/2)# is found by integrating over the limits

#w=int_0^(l/2) (QE_0)/lxdx-int_(l/2)^0 (QE_0)/lydy#
#=>w=(QE_0)/l(|x^2/2|_0^(l/2)+|y^2/2|_0^(l/2))#
#=>w=QE_0l/4#

Equating work done with initial kinetic energy to be given to the mass we get

#1/2Mv_min^2=QE_0l/4#
#=>v_min^2=(QE_0l)/(2M)#

It is given that #(QE_0l)/M=8# in SI unit. Insertig in above equation we get

#v_min^2=8/2#
#:.v_min=2"ms"^-1#