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Question

Consider the figure.
The charge appearing on C2 is

A
E(C3C4C1+C2)
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B
E(C1C2C1+C2)
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C
E(C1C2C3+C4)
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D
E(C3C4C3+C4)
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Solution

The correct option is B E(C1C2C1+C2)
Let us mark the points across the whole combination of capacitors as C and D.From the above fiqure we can see that the voltage across C and D is to E.
VC1=qC1 (voltage across C1)
IIy VC2=qC2
As C1 and C2 are in series, thus
VC1+VC2=E
qC1+qC2=E........(i)
E=C1+C2C1C2q
q=C1C2C1+C2E

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