(20 14? Hefei mode 2) As shown in the figure, in the electric field generated by point charge Q, two positively charged trial charges q 1 and q2 are placed at point A and point B respectively, and the

(20 14? Hefei mode 2) As shown in the figure, in the electric field generated by point charge Q, two positively charged trial charges q 1 and q2 are placed at point A and point B respectively, and the dotted line is A. According to the question, when two positively charged trial charges q 1 and q2 move from point A and point B to point C respectively, the positive work done by the electric field force is equal, and then it is known that there is gravity between Q and the two trial charges, indicating that Q is negatively charged, and the direction of the electric field line points to Q from infinity, so the potential of point A is less than that of point B, so A is wrong;

B, from the analysis of point charge field strength formula E=kQr2, it can be seen that the field strength of point A is greater than that of point B, so B is wrong;

C. From the analysis of the figure, it can be seen that the potential difference between A and C is smaller than that between B and C. In the process of q 1 and q2 moving to C, the work done by the external electric field force is equal. According to the formula W=qU, the charge of q 1 is greater than that of q2, so c is correct.

D. According to the analysis of C, the charge of q 1 is greater than that of q2, and the potential energy at infinity is zero. According to the expression of potential energy: EP=qφ, q 1 and q2 are not equal in potential energy at point C, so d is wrong.

So choose: C.