question_answer

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]

A)                    \[\frac{{{\varepsilon }_{0}}AV}{d}.\frac{2{{\varepsilon }_{0}}AV}{d}\]

B)                    \[\frac{{{\varepsilon }_{0}}AV}{d}.\frac{2{{\varepsilon }_{0}}AV}{d}\] 

C)                    \[\,\frac{{{\varepsilon }_{0}}AV}{d}.\frac{-2{{\varepsilon }_{0}}AV}{d}\]

D)                    \[\frac{-{{\varepsilon }_{0}}AV}{d}.\frac{-2{{\varepsilon }_{0}}AV}{d}\]

Correct Answer: C

Solution :

           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 \[-\frac{2{{\varepsilon }_{0}}A}{d}V\]