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) 4 pF (C) 6.36 pF (D) 12.72 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

4 pF

6.36 pF

12.72 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