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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 energies stored in the combination before and after the introduction of dielectric slabs . |
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Answer» SOLUTION :Total initial energy of the combination `U_i = 1/2 C_i V^(2)` As the KEY K is then thrown OPEN , net charge Q remains constant but voltage V changes to `V_(f) = (V)/(K_(0))` `therefore` Total final energy of the combination `U_(f) = 1/2 C_f V_f^2 = 1/2 (K_(0) . C_(f)) ((V)/(K_(0)))^(2) = (C_(1) * V^(2))/(2 K_(0)) implies (U_(i))/(U_(f)) = K_(0)` |
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