InterviewSolution
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InterviewSolution
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A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants `k_1`, `k_2` and `k_3` as shown. If a isngle dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectic constant k is given by A. `(1)/(k) = (1)/(k_(1)) + (1)/(k_(2)) + (1)/(2k_(3))`B. `(1)/(k) = (1)/(k_(1) + k_(2)) + (1)/(2k_(3))`C. `k = (k_(1)k_(2))/(k_(1) + k_(2)) + 2k_(3)`D. `k = k_(1) + k_(2) + 2k_(3)` |
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Answer» Correct Answer - B `C_(1) = (K_(1)epsilon_(0)(A)/(2))/(((d)/(2))) = (K_(1)epsilon_(0)A)/(d)` `C_(2) = (K_(2)epsilon_(0)(A)/(2))/(((d)/(2))) = (K_(2) epsilon_(0) A)/(d)` and `C_(3) = (K_(3)epsilon_(0)(A)/(2))/(((d)/(2))) = (2K_(3)epsilon_(0)A)/(d)` `= (1)/(C_(eq)) = (1)/(C_(1) + C_(2)) + (1)/(3) = (1)/((epsilon_(0)A)/(d)(K_(1) + K_(2))) + (1)/((epsilon_(0))/(d) xx2K_(3))` `(1)/(C_(eq) = (d)/(epsilon_(0)A) [(1)/(K_(1) + K_(2)) + (1)/(2K_(3))]` `C_(eq) = [(1)/(K_(1) + K_(2)) + (1)/(2K_(3))]^(-1). (epsilon_(0)A)/(d)` So `K_(eq) = [(1)/(K_(1) + K_(2)) + (1)/(2K_(3))]^(-1)` |
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