Find identical capacitor plates, each of area $A$ , are arranged such that adjacent plates are at a distance $^{'} d^{'}$ apart, the plates are connected to a source of emf $V$ as shown in figure. The charge on plate 1 is and that on plate 4 is .

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Solution

We have capacitance,
$C = ε_{o} \frac{A}{d}$ and $C = \frac{Q}{V}$
$⟹ Q = C V$
Hence, charge on Plate 1 is $C = \frac{ϵ_{o} A}{d} V$
and charge on Plate 4 is $C = - \frac{2 ϵ_{o} A}{d} V$

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Find identical capacitor plates, each of area A, are arranged such that adjacent plates are at a distance ′d′ apart, the plates are connected to a source of emf V as shown in figure. The charge on plate 1 is ____________ and that on plate 4 is ____________.

We have capacitance,C=εoAd and C=QV⟹Q=CVHence, charge on Plate 1 is C=ϵoAdVand charge on Plate 4 is C=−2ϵoAdV

Find identical capacitor plates, each of area $A$ , are arranged such that adjacent plates are at a distance $^{'} d^{'}$ apart, the plates are connected to a source of emf $V$ as shown in figure. The charge on plate 1 is and that on plate 4 is .

We have capacitance,
$C = ε_{o} \frac{A}{d}$ and $C = \frac{Q}{V}$
$⟹ Q = C V$
Hence, charge on Plate 1 is $C = \frac{ϵ_{o} A}{d} V$
and charge on Plate 4 is $C = - \frac{2 ϵ_{o} A}{d} V$