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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

Five identica...

Question

Five identical plates, each of area A are arranged such that adjacent plates are at a distance d apart and are numbered as 1, 2, 3, 4 and 5 from left to right. The plates are connected to a battery of voltage V as shown in figure If $Q_{1}$ is the charge on the plate 1 and $Q_{4}$ on plate 4, then,

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Solution

The correct options are
A

D

Plate 1 is connected to the positive terminal of the battery and plate 2 is connected to the
negative terminal. Hence, $Q_{1}$ = CV = $\frac{ϵ_{o} A V}{d}$ plates 1,2,4 and 5 constitute 2 capacitors in parallel;
their combined capacitance is, $C^{1}$ = 2C.

The charge on plate 4 is negative because it is connected to the negative terminal of the battery.

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Five identical plates, each of area A are arranged such that adjacent plates are at a distance d apart and are numbered as 1, 2, 3, 4 and 5 from left to right. The plates are connected to a battery of voltage V as shown in figure If Q1 is the charge on the plate 1 and Q4 on plate 4, then,

Five identical plates, each of area A are arranged such that adjacent plates are at a distance d apart and are numbered as 1, 2, 3, 4 and 5 from left to right. The plates are connected to a battery of voltage V as shown in figure If $Q_{1}$ is the charge on the plate 1 and $Q_{4}$ on plate 4, then,