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20 16 Taizhou sanmao mathematics
A and B deduced from E=Ud and U=QC that the electric field intensity between plates was E=QdC. In the process of moving from position I to position II, the electric field force does negative work on negative charge-qEL and positive work on positive charge QE. 2L, then the work done by the electric field force on the system is W=-qEL+qE? 2L=qEL=qQLdC, so the sum of the potential energy of the two balls reduces qQLdC. So A is right and B is wrong.

When C and D systems are in position I, the total electric potential energy is zero, and when they move to infinity, the total electric potential energy is also zero. Because the sum of the electric potential energy of the two balls decreases qcld in the process of moving from position I to position II and increases qcld in the process of moving from position II to infinity, the system overcomes the electric field force to do work qQLdC. So, C is wrong and D is right.

So choose AD.