Five identical capacitor plates are arranged such that they make capacitors each of $2\  \mu F$. The plates are connected to a source of $emf$ $10\  V$. The charge on plate $C$ is......$ \mu C$
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Here four capacitors are in parallel and charge on each is $q=C V$. As the plate $C$

is the common plate between the two capacitors so its charge will be double, i.e,

$q=2 C V$ and it is conned to positive terminal of cell so its charge should be positive.

thus, $q=2 C V=2 \times 2 \times 10=40 \mu C$

art

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