From the knowledge of geometry, we can know that the orbital radius of particles is R 1=22a? ①,

Particle motion time: t=πR 1v? ②,

Is the particle v=2πa2t? ③

The acceleration of particles in parallel plate capacitors is obtained by kinetic energy theorem: qU= 12mv2,

The solution is: u = mπ 2A24Qt2;

(2) When the particle just comes out of point D, the particle radius is the smallest and the magnetic induction intensity is the largest.

The period of circular motion of particles in magnetic field: T=2πmqB 1, T=2t, and the solution is: B 1=πmqt.

When the particle is tangent to the AC edge, the particle orbital radius is the largest, and the trajectory is shown in the figure.

At this time, the magnetic induction intensity is the smallest. According to mathematical knowledge, R2=2a,

According to Newton's second law: qvB2=mv2R2, the solution is: B2=πm2qt.

The relationship that the magnetic induction intensity should satisfy: π m2qt ≤ b ≤ π Mqt;

(3) After the magnetic field is reversed, the motion trajectory is shown in the figure.

According to geometric knowledge, 3R3=2a-2a,

Trajectory length: l=( 12+ 18)2πR3,

Exercise time: t3=lv, solution: t3=5(2? 1)6t；

Answer: (1) The potential difference between the two plates of the capacitor U = mπ 2A24Qt2;

(2) If the particle does not collide with the M plate and exits from the AD side, the induction intensity should meet the following conditions: π m2qt ≤ b ≤ π Mqt;

(3) The time for a particle to move in a magnetic field is 5(2? 1)6t。