Solution:
Initial charge is \(Q_i = 60 \mu\text{C}\). When filled with a dielectric of constant \(K\), the final charge is \(Q_f = K Q_i = K times 60 \mu\text{C}\). The extra charge flowing is \(\Delta Q = Q_f - Q_i = 120 \mu\text{C}\), which gives \(K = \frac{180}{60} = 3\).
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