Question

A 0.134-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find the time rate of change of electric flux between the plates. V·m/s (b) Find the displacement current between the plates. A

Answer 1

Answer:

a) 1.51Vm/s

b)Id=0.134A

Explanation:

(a) The time rate of change of electric flux between the plates can be computed by using:

$I_c=C\frac{dV}{dt}=\frac{\epsilon_0A}{d}\frac{d(Ed)}{dt}=\epsilon_0\frac{d\Phi_E}{dt}\\\\\frac{\Phi_E}{dt}=\frac{I_c}{\epsilon_0}=\frac{0.134A}{8.85*10^{-12}C/(Nm^2)}=1.51\frac{Vm}{s}$

where Ic is the current of the capacitor, e0 is the dielectric permittivity of vacuum and A is the area of the capacitor.

(b) The displacement current coincides with capacitor current:

Id = 0.134A

hope this helps!!