From the meaning of the question, it can be seen that the electron accelerates in the accelerating electric field, and the speed obtained by the electron is v = 2 qum. After entering the deflection electric field, the electrons do quasi-plano-parabolic motion, that is, parallel to the electric field direction, they do uniform acceleration linear motion with initial velocity of 0, acceleration a=qU, deflection displacement y = 12at2 of dm electrons in the electric field direction, and uniform acceleration linear motion with velocity v = 2 qum in the vertical electric field direction. When t = lv, the particle moves in the electric field.

A. When the sliding contact moves to the right, the acceleration voltage becomes larger, so the velocity v obtained by the electron increases, and the movement time t = lv of the electron in the electric field decreases, so the deflection displacement y = 12at2 of the electron decreases. Because the electrons deflect upward, the position on the screen drops, and A is wrong;

B, when the sliding contact moves to the left, the acceleration voltage becomes smaller, so the velocity v obtained by the electron decreases, and the movement time t = lv of the electron in the electric field becomes larger, so the deflection displacement y = 12at2 of the electron becomes larger, because the electron deflects upward and its position on the screen rises, so b is correct;

C, when the deflection voltage increases, the electric field force on the electron in the electric field increases, that is, the acceleration A of electron deflection increases. Because the velocity V obtained by the electron is unchanged, the time for the electron to move in the electric field is unchanged, A increases the speed at which the electron impacts the screen, and v = V2+(at) 2. Because the acceleration A is unchanged, the speed at which the electron impacts the screen increases, so C is wrong.

D. When the deflection voltage increases, the electric field force on electrons in the electric field increases, that is, the acceleration a of electron deflection increases. Because the acceleration voltage is constant, the speed of electrons entering the electric field is constant, and the time of electrons moving in the electric field is constant, so D is correct.

So choose BD.