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 BE(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