(1) Let the gravity on the ring be Gp, the electric field force be F, and the rope Zhang Liwei be? T, the stress analysis of the ring is shown in the figure, where f = QE = 4×10-5c× 5×104n/c = 2n.

From the equilibrium condition: t? cos37 +N 1=(Gp+F)？ cos53

Alternative data: n1= 0.8n.

Then the pressure of the ring on the rod is n1'= n1= 0.8n.

Let the elastic force of vertical wall A on the rod be n, and the analysis of the rod can be obtained from the moment balance condition:

g？ l2cos 53 = nls in 53+n 1′。 Commander (commanding officer)

Alternative data: n = 2.95n

(2) When the small ring slides down, the speed of the small ring is the maximum when the included angle between the rope and the rod is α. At this time, the resultant force of the small ring along the rod direction is zero (f+g) sIn53 = fsinα, α = 37, that is, when the small ring slides directly below O 1, the speed of the small ring is the maximum, and the small ring slides down S = 0.3m.

According to the kinetic energy theorem: (F+G)s? sin53 -F(s？ sin53 -s？ cos53 )= 12mv2m

Solution, vm = 2.68m m/s

(3) When the electric field force is reversed, the electric field force is just in balance with gravity, and when the small ring slides in the direction perpendicular to the rod, the speed is the highest.

According to the symmetry, the slip of the small ring is s1= 2scos53× cos53 = 2× 0.3× 0.6× 0.6 (m) = 0.216 (m), and the biggest change of the electric potential energy is

△? =Fs 1？ sin 53 = 2×0.2 16×0.8J = 0.3456j

Answer: (1) When the small ring is just loosened, the elastic force of the vertical wall A to the rod is 2.95 N;

(2) The maximum speed that the small ring can reach during sliding is 2.68m/s;

(3) If only the direction of electric field is reversed, other conditions remain unchanged, the maximum change of electric potential energy in the process of circular motion is 0.3456 J. 。