Three large plates are arranged as shown. How much charge will flow through the key $k $ if it is closed?
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Before closing the switch After closing the switch
$q_{1}+q_{2}=2 Q$
$V_{A B}=V_{A C}$ or $\frac{Q_{1}}{C_{1}}=\frac{q_{2}}{C_{2}}$
$\frac{q_{1}}{q_{2}}=\frac{C_{1}}{C_{2}}=2$ or $q_{1}=2 q_{2}$
Solving, we get $q_{1}=\frac{4 Q}{3}, q_{2}=\frac{2 Q}{3}$
Charge flown through $K$ is $-\frac{Q}{2}-\left(-q_{1}\right)--\frac{Q}{2}+\frac{4 Q}{3}=5 Q / 6$
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