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Two parallel plate capacitors of capacitances C_1 and C_2 such that C_2 = 2 C_1 are connected across battery of V volts as shown in the Fig . Initially the key K is kept closed to fully charge the capacitors . The key K is now thrown open and dielectric slabs of dielectric constant K_0 are inserted in the two capacitors to completely fill the gap between the plates . Find the ratio of the net capacitance |
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Answer» Solution :As two capacitors are joined in PARALLEL , the net initial capacitance `C_1 = C_1 + C_2 = C_1 + 2 C_1 = 3 C_1` On introduction of dielectric slabs of dielectric constant `K_0` , new capacitances BECOME `C._1 = K_0 C_1` and `C._2 = K_0 C_2 = 2 K_0 C_1` `therefore` Net final capacitance `C_(f) = C_(1) + C_(2) = K_(0) C_(1) + 2 K_(0) C_(1) = 3K_(0) C_(1) implies (C_(i))/(C_(f)) = (3C_(1))/(3 K_(0) C_(1)) = (1)/(K_0)` |
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