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The expression for the equivalent capacitance of the system shown above is (A is the cross-sectional area of one of the plates)

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Let’s find the capacity of each capacitor
$C_1 = \large\frac{\in_0 A}{d} $
$C_2 = \large\frac{\in_0 A}{2d} $
$C_3 = \large\frac{\in_0 A}{3d} $
The three capacitors are in parallel.
Equivalent capacitance $C = C_1+ C_2+C_3$
$\qquad= \large\frac{\in_0 A}{d} +\frac{\in_0 A}{2d}+\frac{ \in_0 A}{3d}$
$\qquad= \large\frac{\in_0 A}{d} \bigg( 1+ \large\frac{1}{2} +\frac{1}{3} \bigg)$
$\qquad= \large\frac{11}{6} . \bigg( \large\frac{\in_0 A}{d} \bigg)$
answered Jun 17, 2014 by meena.p
 

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