A, according to the slope of the velocity image is equal to the acceleration, it can be seen from the mathematical knowledge that the acceleration decreases during the movement of charged particles from point A to point B, and the electric field force on charged particles decreases from F=ma, and the electric field intensity E decreases from F=qE, that is, the field strength at A must be greater than that at B. Therefore, A is correct.

B, because the electric properties of charged particles are unknown, it is impossible to judge the direction of electric field, so it is impossible to judge the level of potential. So, B is wrong.

C, D As can be seen from the figure, when the velocity of charged particles increases and the kinetic energy increases, it can be known from the law of conservation of energy that the electric potential energy decreases and the electric field force will inevitably do positive work on the particles. Therefore, the CD is correct.

So: ACD.