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 \[{{q}_{2}}=-2CV=\frac{-2{{\in }_{0}}AV}{d}.\]You need to login to perform this action.
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