A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. (a) what is the displacement current densityin the air space between the plates? (b) What is the rate at which the electric field between the plates is changing? (c) What is the induced magnetic field between the plates at a distance of 2.00 cm from the axis? (d) At 1.00 cm from the axis?

To find the displacement current density in the sir space between the plates, according to the equations:

$$\begin{gathered} i_C=\frac{dq}{dt}\frac{d\varphi E}{dt} \\ {i}_D=\frac{d\varphi E}{dt} \end{gathered}$$

Therefore, we can see that the condition and the displacement currents are equal, i.e.$i_C=i_D$

Hence the displacement current density is:

$$\begin{gathered} j_D=\frac{i_D}{A}= \\ \frac{i_C}{A}=\frac{0.520A}{\pi r^2} \\ =\frac{0.520A}{\pi (0.0400m^2)} \\ =103A/m^2 \end{gathered}$$

Therefore, the displacement current density is$103A/m^2$

To calculate the rate at which electric field between the plates is changing .

The displacement current is:

$$\begin{gathered} i_D={\mathrm{E}}_0\frac{d\varphi E}{dt} \\ =\mathrm{E}A\frac{dE}{dt}{j}_D \\ =\frac{i_D}{A}={\mathrm{E}}_0\frac{dE}{dt}\frac{dE}{dt} \\ =\frac{{\mathrm{j}}_{\mathrm{D}}}{{\mathrm{E}}_0}=\frac{103}{{\displaystyle 8.854\times {10}^{-12}}} \\ =1.16\times {10}^{-12}\mathrm{V}/\mathrm{m}.\mathrm{s} \end{gathered}$$

Therefore, the rate of change is$1.16\times {10}^{-12}\mathrm{V}/\mathrm{m}.\mathrm{s}$

For the plates kept at a distance of from the axis.

$\oint \overrightarrow{B}·\overrightarrow{dl}={\mu }_0(i_C+i_D)$

Since no conduction current flows through the space, and the displacement current is enclosed by the path is $i_D=j_D\pi r^2$.

$B\left(2\pi r\right)={\mu }_0\left(j_D\pi r^2\right)$

Or,

$$\begin{gathered} B=\frac{1}{2}{\mu }_0{j}_{D}r \\ =\frac{1}{2}(4\pi \times {10}^{-}7T·m/A)(103A/m^2)(0.0200m) \\ =1.30\times {{10}^{-}}^{6}T \end{gathered}$$

For the distance of 1.00cm

$$\begin{gathered} B=\frac{1}{2}{\mu }_0{j}_{D}r \\ =\frac{1}{2}(4\pi \times {10}^{-}7T·m/A)(103A/m^2)(0.0100m) \\ =6.50\times {{10}^{-}}^{6}T \end{gathered}$$

Therefore, the magnetic field at a plate separation of is$6.50\times {{10}^{-}}^{6}T$