Question

consider a parallel plate capacitor with circular plates of area A. at an instant, the charge on the capacitor is dq/dt.the displacement current through a circular region of area A/2 parallel to the plates, between them is
a. 1/4 dq/dt
b. 2 dq/dt
c. 1/2 dq/dt
d. dq/dt​

Answer 1

Explanation:

A parallel - plate capacitor is being charged.Show that the displacement current across an area in the region between the plates and parallel to it is equal to the conduction current in the connecting wires.

Answer 2

Answer:

The displacement current through a circular region of area A/2 parallel to the plates, between them is $\frac{1}{2}\frac{dq}{dt}$.i.e.option (c).

Explanation:

The displacement current in terms of dq/dt is given as,

$\frac{dq}{dt}=\epsilon_0A\frac{dE}{dt}$        (1)             ($I=\frac{dq}{dt}$)

Where,

$\frac{dq}{dt}$= displacement current

∈₀=permitivity of the free space

A=area of the plates of the capacitor

E=electric field present between the plates of the capacitor

As per the question, the displacement current through a circular region of area A/2 parallel to the plates is given as,

$\frac{dq'}{dt}=\epsilon_0\frac{A}{2} \frac{dE}{dt}$     (2)

By writing equation (2) in terms of equation (1) we get;

$\frac{dq'}{dt}=\frac{1}{2}\frac{dq}{dt}$        (3)

Hence, the displacement current through a circular region of area A/2 parallel to the plates, between them is $\frac{1}{2}\frac{dq}{dt}$.i.e.option (c).