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Three plates of common surface area `A` are connected as shown. The effective capacitance will be A. `(epsilon_(0)A)/(d)`B. `(3epsilon_(0)A)/(d)`C. `(3)/(2)(epsilon_(0)A)/(d)`D. `(2epsilon_(0)A)/(d)` |
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Answer» Correct Answer - D The given circuit is equivalent to parallel combination of two identical capacitors, each having capacitance `C = (epsilon_(0)A)/(d)`. Hence `C_(eq) = 2C = (2epsilon_(0)A)/(d)` |
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