A gang capacitor is formed by interlocking nine plates with each other. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plate is 5 cm 2. The capacity of the unit is

Question

A gang capacitor is formed by interlocking nine plates with each other. The distance between the consecutive plates is 0.885 cm and the overlapping area of the plate is 5 cm 2. 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

According to the question, 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 nine plates with each other. The distance between the consecutive plates is 0.885cm and the overlapping area of the plate is 5cm2. The capacity of the unit is

1.06 pF

4 pF

6.36 pF

12.72 pF

According to the question, 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 nine plates with each other. The distance between the consecutive plates is $0.885 c m$ and the overlapping area of the plate is $5 c m^{2}$ . The capacity of the unit is