Q 12-02-185JEE MainJEE Main 2019 (11 Jan, Shift 1)Medium
In the figure shown below, the charge on the left plate of the $10\ \mu$F capacitor is $-30\ \mu$C. The charge on the right plate of the $6\ \mu$F capacitor is:
Answer: (D) $+18\ \mu$C
The right plate of the $10\ \mu$F capacitor carries $+30\ \mu$C. The isolated node between it and the parallel pair has zero net charge, so the left plates of the $6\ \mu$F and $4\ \mu$F capacitors together carry $-30\ \mu$C.
Both have the same voltage, so the charge divides in the ratio $6 : 4$:
$$q_6 = \frac{6}{10}\times30 = 18\ \mu\text{C}$$
Left plate of the $6\ \mu$F: $-18\ \mu$C, so its right plate carries $+18\ \mu$C.
Solution by Sreeraj P, M.Sc Physics