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The capacitance of a parallel plate capacitor with plate area `A` and separation `d` is `C`. The space between the plates in filled with two wedges of dielectric constants `K_(1)` and `K_(2)` respectively. Find the capacitance of resulting capacitor.

Answer» Correct Answer - `(CK_(1)K_(2))/(K_(2)-K_(1))ln`(K_(2))/(K_(1))`; where `C=(epsilon_(0)A)/(d)`.
`(CK_(1)K_(2))/(K_(2)-K_(2))ln`(K_(2))/(K_(1)), where C=(epsilon_(0)A)/(d)`.


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