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Five identical plates each of area A are joined as shown in the figure. The distance between the plates is d. The plates are connected to a potential difference of V volts . The charge on plates 1 and 4 will be

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$\frac{2 ε_{0} A V}{d}, - \frac{2 ε_{0} A V}{d}$

$\frac{ε_{0} A V}{d}, - \frac{2 ε_{0} A V}{d}$

$\frac{{{\varepsilon _0}AV}}{d}.\frac{{2{\varepsilon _0}AV}}{d}$

$\frac{{ - {\varepsilon _0}AV}}{d}.\frac{{ - 2{\varepsilon _0}AV}}{d}$

answer is C.

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Detailed Solution

The given circuit can be redrawn as follows. All capacitors are identical and each having capacitance $C = \frac{{{\varepsilon _0}A}}{d}$
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|Charge on each capacitor| = |Charge on each plate| $= \frac{{{\varepsilon _0}A}}{d}V$

Plate 1 is connected with positive terminal of battery so charge on it will be $+ \frac{{{\varepsilon _0}A}}{d}.V$

Plate 4 comes twice and it is connected with negative terminal of battery, so charge on plate 4 will be $- \frac{{2{\varepsilon _0}A}}{d}V$

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