A plane parallel plate capacitor has plates of 10 cm² area that are 1.0 mm apart. At an instant when charge is being accumulated on the plates at a rate of 12 nC/s, the displacement current between the plates is A) 1.06 × 10−¹⁶ A. B) 1.2 × 10−⁸ A. C) 8.85 × 10−⁹ A. D) 1.00 A. E) 1.36 A.

Question

Quizplus · Accepted Answer

The displacement current between the plates of a capacitor is given by the expression $I_D = \epsilon_0\frac{d\Phi_E}{dt}$, where $\epsilon_0$ is the permittivity of free space and $\frac{d\Phi_E}{dt}$ is the rate of change of electric flux through the surface bounded by the plates.

The electric field between the plates of the capacitor due to a charge $Q$ on each plate is given by $E = \frac{Q}{\epsilon_0 A}$, where $A$ is the area of each plate. Since the plates are parallel, the electric flux through the surface bounded by the plates is $\Phi_E = EA$. Therefore, the rate of change of electric flux is $\frac{d\Phi_E}{dt} = A\frac{dE}{dt}$.

The rate of accumulation of charge on each plate is given by $\frac{dQ}{dt} = 12$ nC/s. Therefore, the electric field between the plates is changing at a rate of $\frac{dE}{dt} = \frac{\frac{dQ}{dt}}{\epsilon_0 A} = \frac{12 	imes 10^{-9}}{\epsilon_0(0.1^2)}$ V/m.

Substituting this value in the expression for displacement current, we get $I_D = \epsilon_0 A\frac{dE}{dt} = \epsilon_0(0.1^2)\frac{12	imes10^{-9}}{\epsilon_0(0.1^2)} = 12	imes10^{-9}$ A = 1.2 $\mu$A. Therefore, the best choice is B.

A Plane Parallel Plate Capacitor Has Plates of 10 Cm2

A Plane Parallel Plate Capacitor Has Plates of 10 Cm²