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Q.

The network shown in the diagram, shows two identical parallel plate capacitors each of capacitance 2μF. Now separation between the plates of (1) is doubled and that between the plates of (2) is halved. Then in the process charge flown through the battery is

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a

4 μC

b

2 μC

c

6 μC

d

0.5 μC

answer is B.

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Detailed Solution

Initially C1=C2=2μF⇒Ce=2 x 22+2μF=1μF∴Qi=Initial charge on the system of capacitors =1 x 10 μC=10 μCFinally , C1'=22 μF=1 μF and C2'=2 x 2 μF=4 μF ∵C=ε0Ad∴Ce'=1 x 41+4μF=45μF⇒Qf=45x10μC=8μCCharge flown through the battery =Qi-Qf=(10-8)μC=2μC

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