A parallel plate capacitor is charged to 100 × 10-6 C. Due to radiations falling, from a radiating source the plate loses charge at the rate of 2 × 10-7 Cs -1. The magnitude of displacement current is

Question

A parallel plate capacitor is charged to 100 × 10-6 C. Due to radiations falling, from a radiating source the plate loses charge at the rate of 2 × 10-7 Cs -1. The magnitude of displacement current is (A) 10-6 A (B) 10-4 A (C) 2 × 10-7 A (D) 2 × 10-7 mCs -1

Tardigrade Admin · Accepted Answer

Magnitude of displacement current is given by Id=IC=|(d q/d t)|=2 × 10-7 Cs -1=2 × 10-7 A

Q. A parallel plate capacitor is charged to $100 \times 1 0^{- 6} C$ . Due to radiations falling, from a radiating source the plate loses charge at the rate of $2 \times 1 0^{- 7} C s^{- 1}$ . The magnitude of displacement current is

$1 0^{- 6} A$

$1 0^{- 4} A$

$2 \times 1 0^{- 7} A$

$2 \times 1 0^{- 7} m C s^{- 1}$

Magnitude of displacement current is given by
$I_{d} = I_{C} = ∣ ∣ \frac{d q}{d t} ∣ ∣ = 2 \times 1 0^{- 7} C s^{- 1} = 2 \times 1 0^{- 7} A$

Q. A parallel plate capacitor is charged to 100×10−6C. Due to radiations falling, from a radiating source the plate loses charge at the rate of 2×10−7Cs−1. The magnitude of displacement current is

10−6A

10−4A

2×10−7A

2×10−7mCs−1