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Physics Question on electrostatic potential and capacitance

A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm0.885\, cm and the overlapping area of the plates is 5 cm25 \,cm^2. The capacity of the unit is

A

1.06 PF

B

12.72 PF

C

4 PF

D

6.36 PF

Answer

4 PF

Explanation

Solution

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,
C=ε0AdC =\frac{\varepsilon_{0} A}{d}
=8.854×10−12×5×10−40.885×10−2=0.5 PF=\frac{8.854 \times 10^{-12} \times 5 \times 10^{-4}}{0.885 \times 10^{-2}}=0.5\, PF
Hence, the capacity of 8 capacitors
=8C=8×0.5=4 pF=8 C=8 \times 0.5=4 \,pF