Question

As a parallel-plate capacitor with circular plates 24 cm in diameter is being charged, the current density of the displacement current in the region between the plates is uniform and has a magnitude of 20 A/m2.

(a) Calculate the magnitude B of the magnetic field at a distance r = 87 mm from the axis of symmetry of this region.

(b) Calculate dE/dt in this region.

Answer 1

Answer:

Explanation:

a.)

The magnitude field

$B=\frac{\mu _0I_{enclosed}}{2\pi r}\\\\=\frac{\mu _0(J_ar^2)}{2\pi r}\\\\=\frac{1}{2}(\mu _0J_ar)\\\\(0.5)(4\pi \times 10^{-7})(20A/m^2)(87\times 10^{-3})\\\\=1.0933\times 10^{-6}T$

b.)

The displacement current

$i_d=\epsilon_0A\frac{dE}{dt}$

then $\frac{dE}{dt}=\frac{i_d}{\epsilon_0A}\\\\=\frac{J_d}{\epsilon_0}\\\\=\frac{20}{8.85\times 10^{-12}}\\\\=2.26\times 10^{12}V/ms$