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
$n C$
$(n + 1) C$
$(n - 1) C$
$\frac{n}{C}$

Question

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
$n C$
$(n + 1) C$
$(n - 1) C$
$\frac{n}{C}$

Toppr Admin · Accepted Answer

The correct option is D -n-x2212-1-CIf n equally spaced plates are stacked alternatively then its diagramatic representation can be shown as follows-xA0-  Upon furthur simplication- the above figure can be rearranged as We can see that the number of capacitors formed due to 4 plates alternatively are 3- Therefore- if n number of plates are to be stacked alternatively the the number of capacitors formed in parallel will be -n-x2212-1- -x27F9-Ceq-n-x2212-1-C

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 isnC(n+1)C(n−1)CnC

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
$n C$
$(n + 1) C$
$(n - 1) C$
$\frac{n}{C}$