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 $5 \times 10^{13} Vms ^{-1}$.

• A. $13.8 A$
• B. $12.6 A$
• C. $13.9 A$
• D. $10.5 A$

Answer 1

Answer: (C)

Here, cross-section area of a capacitor i.e.,
$A=\pi R^{2}=\pi(0.1 m )^{2}=3.14 \times 10^{-2} m ^{2}$
i.e., $\frac{d E}{d t}=5 \times 10^{13} Vms ^{-1}$
Thus, displacement current
$ { i.e., } i_{d} =\varepsilon_{0} A \frac{d E}{d t} $
$=\left(8.85 \times 10^{-12} C ^{2} Nm ^{-1}\right)\left(3.14 \times 10^{-2} m ^{2}\right) $
$\left(5 \times 10^{13} Vms ^{-1}\right) =13.9 A$