According to mathematical knowledge, the tangent slope of Ep-x image is equal to △Ep△x, and the tangent slope at x 1 is zero, so the electric field intensity at x 1 is zero, so A is correct.
B, according to the relationship between electric potential and electric potential: Ep=qφ, because particles are negatively charged, and q < 0, we know that the greater the electric potential, the lower the electric potential where particles are located, so there is: φ 1 > φ 2 > φ 3. Therefore, B is correct.
C, d, as can be seen from the figure, the tangent slope of the image at the section of 0 ~ X 1 continuously decreases. According to the above formula, when the field strength decreases, the electric field force acting on the particles decreases, the acceleration decreases, and the particles move at a non-uniform speed. The tangent slope of the image on the cross section X 1 ~ X2 increases, and the electric field force acting on the particles increases, and the particles move at a non-uniform speed. The gradient of X2 ~ X3 section remains unchanged.
So: ABD.