Two identical conducting very large plates P1 and P2having charges +4Q and +6Q are placed very closed to each other at separation d. The plate area of either face of the plates is A. The potential difference between plates P1 and P2 is

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VP1−VP2=QdAε0

VP1−VP2=−QdAε0

VP1−VP2=5QdAε0

VP1−VP2=−5QdAε0

answer is B.

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Detailed Solution

Potential difference between the plates isVP1−VP2=−Q(ε0A/d)=−Qdε0A

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