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Question

# Five identical capacitor plates are arranged such that they make four capacitors each of 2μF. The plates are connected to a source of emf 10 V. The charge on plate C is :+40μF+60μF+20μF+80μF

A
+80μF
B
+60μF
C
+20μF
D
+40μF
Solution
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#### Here four capacitors are in parallel and charge on each is q=CV.As the plate C is the common plate between the two capacitors so its charge will be double ,i.e,q=2CV and it is conned to positive terminal of cell so its charge should be positive.thus, q=2CV=2×2×10=40μC

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