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Huainan dual-mode mathematics answer
The potential at point a is φA=-kQR+kQd? R=-kQ(d? 2R)R(d? r); The potential at point C is φC=-kQR+kQd+R=-kQdR(d+R).

Then the potential difference between a and c is UAC=φA-φC=-kQ(d? 2R)R(d? R)-(-kQdR(d+R))=2kQRd2? R2,

Protons move from a to c, and the electric field force does work as WAC=eUAC=2kQeRd2? R2 is positive work, so the proton potential energy decreases by 2kQeRd2? R2, so A is correct.

So choose: a.