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    				| 1. | For the congiuration of mede of permittiveties `epslon_(0), epsilon,` and `epsilon_(0)` between parallel plates each of area `A`, as shown in the equivalent capacitance is . .A. `epsilon_(0)A/d`B. `epsilonepsilon_(0)A//d`C. ` (epsilonepsilon_(0)A)/(d(epsilon+epsilon_(0)))`D. `(epsilonepsilon_(0)A)/((2epsilon+epsilon_(0))d)`. | 
| Answer» Correct Answer - D `C_(eq)=(epsilon_(0))/(d/K_(1)+d/K_(2)+d/K_(3)` Hence, `K_(1)=K_(3)=1, K_(2)epsilon_(0)`. | |