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 volts. The charge on plates 1 and 4 will be

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

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 :

[c] The charge on the plates are shown in figure. The capacitance between neighboring Two plates, \[C=\frac{{{\in }_{0}}A}{d}\]charge on plate 1, \[{{q}_{1}}=+CV=\frac{{{\in }_{0}}AV}{d},\] and \[{{q}_{2}}=-2CV=\frac{-2{{\in }_{0}}AV}{d}.\]