Five identical capacitor plates are arranged such that they make four capacitors each of $2 μ F$ . The plates are connected to a source of emf $10 V$ . The charge on plate $C$ is :

Question

Five identical capacitor plates are arranged such that they four capacitors each of 2 mu F- The plates are connected to a of emf 10 V- The charge on plate C is -40 mu F-60 mu F-20 mu F-80 mu F

Toppr Admin · Accepted Answer

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, $q = 2 C V = 2 \times 2 \times 10 = 40 μ C$

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

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

Five identical capacitor plates are arranged such that they make four capacitors each of $2 μ F$ . The plates are connected to a source of emf $10 V$ . The charge on plate $C$ is :

$+ 40 μ F$
$+ 60 μ F$
$+ 20 μ F$
$+ 80 μ F$