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
IIT 1984, Diffcult
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(c)The given circuit can be redrawn as follows. All capacitors are identical and each having capacitance $C = \frac{{{\varepsilon _0}A}}{d}$
|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$
|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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