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 \[-\frac{2{{\varepsilon }_{0}}A}{d}V\]You need to login to perform this action.
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