In the given circuit, a charge of +80 μC is given to the upper plate of the 4 μF capacitor. Then in the steady state, the charge on the upper plate of the 3 μF capacitor is
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IIT-JEE 2012 · Paper 2, Physics question 4
Answer and worked solution
Official answer: C
Charge distribution. The 4 μF capacitor's upper plate is isolated with +80 μC, so in the steady state its lower plate carries −80 μC. The node joining it to the 2 μF and 3 μF capacitors is isolated and neutral, so the upper plates of those two carry +80 μC in total.
Split. The 2 μF and 3 μF capacitors are in parallel between that node and earth, so the charge divides in proportion to capacitance:
(C)