Question

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?

Short answer

Answer:

(a) When the circular plate has a radius of 4.00cm and at a particular instant the conduction in the wire is0.520 A the displacement current density in the air space between the plates is$103A/m^3$

(b) The electric field between the plates is changing at a rate of$1.16\times {{10}^1}^{3}V/m·s$

(c) The induced magnetic field between the plates at a distance of 2.00cm from the axis is$1.30\times {{10}^{-}}^{6}T$

(d) The induced magnetic field between the plates at a distance of from the axis is$6.50\times {{10}^{-}}^{7}T$

Step-by-step solution

Calculating the current displacement density

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$103A/m^2$

Rate of change of electric field

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$1.16\times {10}^{-12}\mathrm{V}/\mathrm{m}.\mathrm{s}$

Calculating the induced magnetic field in the given distance

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$6.50\times {{10}^{-}}^{6}T$