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Five identical capacitor plates, each of...

Five identical capacitor plates, each of area A, are arranged such that adjacent plates are at distance d apart. The plates are connected to a source of emf V as shown in figure.

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When the plates are alternately connected the capacitors are treated in parallel . In the diagram we have four capacitors so total capacitance is `(4epsilon_(0)A)/(d)` . Now total charge on positive or negative plates is `(4epsilon_(0)AV)/(d)` . Remember that these positive or negative charges are distributed in four layers as shown below. On the outer plates no layer has been shown because no charge is there to bind it . On the plate 2,3,4 double layer charges have been shown because binding opposite charges are on both thes sides of these plates

You know that as per the conservation of charge principle positive and negative charges will be equal in magnitude . Now as there are four layers of charges and the total charge is `(4epsilon_(0)AV)/(d)` the charge on single layer is `(epsilon_(0)AV)/(d)`. So on plate 1 we have single layer positive charge equal to `(epsilon_(0)AV)/(d)` . On plate 4 we have double layer of negative charges and its value is `(-2epsilon_(0)AV)/(d)`.
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AAKASH INSTITUTE ENGLISH-ELECTROSTATIC POTENTIAL AND CAPACITANCE -EXAMPLE
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