How do you solve?

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2 Answers
Oct 11, 2017

See below.

Explanation:

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)

Oct 13, 2017

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