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A parallel plate capacitor is made by stacking $n$ equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is

$\begin {array} {1 1} (A)\;(n-1)C & \quad (B)\;(n+1)C \\ (C)\;C & \quad (D)\;nC \end {array}$

 

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$C_{eq}=(n-1)c \: ( \because$ all capacitors are in parallel )
Ans : (A)
answered Feb 15, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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