Solution: Solution: (1) The charge first passes through the point P (2,0) and enters the I-th quadrant through the X axis. After t=2× 10-4s, it returns to the point P (2,0) at the same speed and in the opposite direction, indicating that the charge first decelerates uniformly in the first quadrant, and then in the opposite direction.

(2) The trajectory of charge is shown in the figure. There is a uniform magnetic field in the fourth quadrant, and the charge moves in a uniform circle, which is obtained by Newton's second law:

qvB=mv2R，

According to the knowledge of geometry, θ = 45× 2 = 90,

2Rcos45 =。 OP=2，

Solution: R= 1m, b = 0.02t t.

The charge moves at a uniform speed in the electric field, which is obtained by Newton's second law: a=qEm,

According to the speed formula of uniform variable speed motion, 0=v0-at2,

Simultaneous and alternative data: e = 200 n/c;

(3) The position where the charge passes through the X axis for the third time: X = XP+2COS = 22m；;

(4) After passing through the X axis for the third time, the charge moves like a flat throw, and the initial velocity direction is R=v0t.

Direction of electric field force: R= 12a2t22,

According to Newton's second law: a2=qE'm,

Simultaneous and alternative data: e ′ =1.6×104n/c;

Answer: (1) The charge is negatively charged, and there is a uniform electric field in the first quadrant. Direction: the positive direction of the X axis is inclined upward by 45;

(2) The magnetic induction intensity is 0.02T, and the electric field intensity is 200 n/c;

(3) The position where the charge passes through the X axis for the third time is 22m；;

(4) The changed electric field intensity is1.6×104 N/c. 。