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