A parallel-plate capacitor consists of two circular plates with radius R=10 cm separated by distance d=0.5 mm. The capacitor is being changed at a uniform rate by applying a changing potential difference between the two plates. Calculate the displacement current for the capacitor. Assume that the electric field is due to the displacement current only and rate at which the electric field between the plates changes is 5 × 1013 Vms -1.

Question

A parallel-plate capacitor consists of two circular plates with radius R=10 cm separated by distance d=0.5 mm. The capacitor is being changed at a uniform rate by applying a changing potential difference between the two plates. Calculate the displacement current for the capacitor. Assume that the electric field is due to the displacement current only and rate at which the electric field between the plates changes is 5 × 1013 Vms -1. (A) 13.8 A (B) 12.6 A (C) 13.9 A (D) 10.5 A

Tardigrade Admin · Accepted Answer

Here, cross-section area of a capacitor i.e.,

A = π R^{2} = π (0.1 m)^{2} = 3.14 \times 1 0^{- 2} m^{2}

i.e.,

\frac{d E}{d t} = 5 \times 1 0^{13} Vm s^{- 1}

Thus, displacement current

i . e ., i_{d} = ε_{0} A \frac{d E}{d t}

= (8.85 \times 1 0^{- 12} C^{2} N m^{- 1}) (3.14 \times 1 0^{- 2} m^{2})

(5 \times 1 0^{13} Vm s^{- 1}) = 13.9 A

$13.8 A$

$12.6 A$

$13.9 A$

$10.5 A$

13.8A

12.6A

13.9A

10.5A

Here, cross-section area of a capacitor i.e., A=πR2=π(0.1m)2=3.14×10−2m2 i.e., dtdE​=5×1013Vms−1 Thus, displacement current i.e.,id​=ε0​AdtdE​ =(8.85×10−12C2Nm−1)(3.14×10−2m2) (5×1013Vms−1)=13.9A

$13.8 A$

$12.6 A$

$13.9 A$

$10.5 A$