JEE Advanced · Physics · 18. Capacitance
A circuit is connected as shown in the figure with the switch \(S\) open. When the switch is closed, the total amount of charge that flows from \(Y\) to \(X\) is

- A zero
- B \(54 \mu \mathrm{C}\)
- C \(27 \mu \mathrm{C}\)
- D \(81 \mu \mathrm{C}\)
Answer & Solution
Correct Answer
(C) \(27 \mu \mathrm{C}\)
Step-by-step Solution
Detailed explanation
From \(Y\) to \(X\) charge flows to plates \(a\) and \(b\).
\(
\left(q_0+q_b\right)_i=0,\left(q_0+q_b\right)_f=27 \mu \mathrm{C}
\)


\(\therefore 27 \mu \mathrm{C}\) charge flows from \(Y\) to \(X\).
\(\therefore\) Correct option is (c)
\(
\left(q_0+q_b\right)_i=0,\left(q_0+q_b\right)_f=27 \mu \mathrm{C}
\)


\(\therefore 27 \mu \mathrm{C}\) charge flows from \(Y\) to \(X\).
\(\therefore\) Correct option is (c)
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