A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 cm2. The capacity of the unit is

Question

A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plates is 5 cm2. The capacity of the unit is (A) 1.06 PF (B) 12.72 PF (C) 4 PF (D) 6.36 PF

Tardigrade Admin · Accepted Answer

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor,

C = \frac{ε _{0} A}{d}

= \frac{8.854 \times 1 0 ^{- 12} \times 5 \times 1 0 ^{- 4}}{0.885 \times 1 0 ^{- 2}} = 0.5 PF

Hence, the capacity of 8 capacitors

= 8 C = 8 \times 0.5 = 4 pF

Q. A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is 0.885cm and the overlapping area of the plates is 5cm2. The capacity of the unit is

1.06 PF

12.72 PF

4 PF

6.36 PF

The given arrangement of nine plates is equivalent to the parallel combination of 8 capacitors. The capacity of each capacitor, C=dε0​A​ =0.885×10−28.854×10−12×5×10−4​=0.5PF Hence, the capacity of 8 capacitors =8C=8×0.5=4pF

Q. A gang capacitor is formed by interlocking a number of plates as shown in figure. The distance between the consecutive plates is $0.885 c m$ and the overlapping area of the plates is $5 c m^{2}$ . The capacity of the unit is