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Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K₁, K₂ and K₃. The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be (E₁refers to capacitor (I) and E₂to capacitor (II)) :options |
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Answer» E1/E2 = 9k1k2k3/(k1+K2+k3)[k2k3+k3k1+k1k2] Option B is correct •When a dielectric of dielectric CONSTANT K is inserted between two plates of CAPACITOR then : C' = KC ______(1) • I) Arrangement will behave as series combination of the capacitor each with capacitance kAE/(d/3) i.e. 3kAE/d where , k is dielectric constant HENCE net capacitance will be 1/C = 1/C1 + 1/C2 + 1/C3 1/C = 1/(3k1AE/d) + 1/(3k2AE/d) + 1/(3k3AE/d) 1/C = d/3AE [ 1/k1 + 1/K2 + 1/K3] 1/C = d[k2k3+k3k1+k1k2]/3AE.[k1k2k3] C = 3AE[k1k2k3]/d[k2k3+k3k1+k1k2] •E1 = CV²/2 •E1 = 3AE[k1k2k3]V²/2d[k2k3+k3k1+k1k2] _______(2) •Similarly II) Arrangement will behave as parallel combination of the capacitor each with capacitance k(A/3)E/d i.e. kAE/3d where , k is dielectric constant Hence net capacitance will be C = C1 + C2 + C3 C = k1AE/3d + k2AE/3d + k3AE/3d C = AE( k1 + k2 + k3 )/3d E = CV²/2 E2 = AE( k1 + k2 + k3 )V²/3(2)d _______(3) Dividing (2) &(3) •E1/E2 = {3AE[k1k2k3]V²/2d[k2k3+k3k1+k1k2]}/ AE( k1 + k2 + k3 )V²/3(2)d •E1/E2 = 9k1k2k3/(k1+k2+k3)[k2k3+k3k1+k1k2] |
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