Q7.35P - Introduction to Electrodynamics

Question

Question: The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (Fig. 7.46a). Again, the current  I is constant, the radius of the capacitor is a, and the separation of the plates is  w << a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0. 

(a) Find the electric field between the plates, as a function of t. 

(b) Find the displacement current through a circle of radius   in the plane mid-way between the plates. Using this circle as your "Amperian loop," and the flat surface that spans it, find the magnetic field at a distance s from the axis.

                             

                                                            Figure 7.46

(c) Repeat part (b), but this time uses the cylindrical surface in Fig. 7.46(b), which is open at the right end and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to I_enc. 

Step-by-Step Solution

Answer

Answer

(a) The value of the electric field between the plates, as a function of  is $E=\frac{It}{S{\epsilon }_0}$.

(b) 

The value of displacement current $I_d=EdS$  .

The value of magnetic field at a distance  s from the axis is  $\overrightarrow{B}=\frac{{\mu }_{0}I}{2}\frac{s}{A}\hat{\varphi }$.

(c) The value of magnetic field in the same direction as the case b is $\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi }\frac{s}{a^2}$  .

Step 1: Write the given data from the question.

Consider that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at  t = 0.

Step 2: Determine the formula of electric field between the plates, as a function of t.

Write the formula of electric field between the plates, as a function of .

                                                  $\mathbf{E}=\frac{\sigma \left(\mathbf{t}\right)}{{\mathbf{\epsilon }}_{\mathbf{0}}}$                                              …… (1)

Here, $\sigma \left(\mathbf{t}\right)$  is charge on the plate and  ${\mathbf{\epsilon }}_{\mathbf{0}}$ is relative permittivity.

Write the formula of magnetic field at a distance s  from the axis.

               $\mathbf{B}=\frac{{\mu }_{\mathbf{0}}{\mathbf{I}}_{\mathbf{d}}}{\mathbf{2}\pi s}$                                                                                   …… (2)

Here,  ${\mu }_{\mathbf{0}}$ is permeability,  $I_d$ is displacement current and  s is the surface of the plates.

Write the formula of magnetic field in the same direction as the case b.

                          $\mathbf{B}=\frac{{\mu }_{\mathbf{0}}\mathbf{I}}{2\pi \mathbf{s}}\left[\mathbf{1}-\frac{\mathbf{I}\left(\mathbf{s}\right)}{\mathbf{I}}\right]$                                                        …… (3)

Here,  ${\mu }_{\mathbf{0}}$ is permeability, I  is current,  I_S is current passing through curve.

Step 3: (a) Determine the electric field between the plates.

Determine the charge on the plate is:

  $$\begin{gathered} Q\left(t\right)=\sigma \left(t\right)S \\ =It \end{gathered}$$

Since the charge density grows linearly with time and is homogeneous throughout the plate. The electric field is thus:

Determine the electric field between the plates, as a function of t.

Substitute  $\frac{Q\left(t\right)}{S}$ for $\sigma \left(t\right)$  into equation (1).

  $$\begin{gathered} E=\frac{Q\left(t\right)}{{\epsilon }_{0}S} \\ =\frac{It}{S{\epsilon }_0} \end{gathered}$$

Here, S  is the surface of the plates.

Step 4: (b) Determine the displacement current and magnetic field at a distance s from the axis.

The magnetic field will curve around the current if it flows down the z-axis, using Ampere's equation with the displacement current term.

Determine the displacement current.

 $I_d=EdS$ 

Determine the magnetic field at a distance   from the axis.

Substitute  EDS for  I_d into equation (2).

  $$\begin{gathered} B=\frac{{\mu }_0{\epsilon }_0}{2\pi s}\frac{d}{dt}{\int }_{S}EdS \\ =\frac{{\mu }_0\left(s^2\pi \right)}{2\pi s}\frac{I}{\pi a^2} \\ =\frac{{\mu }_{0}I}{2}\frac{s}{A}\hat{\varphi } \end{gathered}$$

Whereas, the integration loop was positioned between the capacitor plates as a circle whose radius was coaxial with the plates.

Step 5: (c) Determine the magnetic field in the same direction as the case b.

Using a new bounding surface but the same Amperian loop once more, the magnetic field is:

  $2\pi sB={\mu }_0{I}_{enc}$

There will be two contributions from "regular" current but there is contribution from the displacement current. Given that the surface's normal points inwards, one is on surface S₁   (positive), and the other is at curve C₁  (negative).

  $$\begin{gathered} 2\pi sB={\mu }_0{I}_{enc} \\ 2\pi sB={\mu }_{0}I-{\mu }_{0}I\left(s\right) \\ B=\frac{{\mu }_{0}I}{2\pi s}\left[1-\frac{I\left(s\right)}{I}\right] \end{gathered}$$

The charge density must only be a function of time for the current I(s) to travel through curve . Let's say that all we search for is the charge density on the circle enclosed by curve :

  $$\begin{gathered} \sigma \left(t\right)=\frac{Q\left(s,t\right)}{s^2\pi } \\ =\frac{1}{s^2\pi }\left[I-I\left(s\right)\right]t \end{gathered}$$

For   to only be a function of time the term $I-I\left(s\right)$  needs to be proportional to ;

  $I-I\left(s\right)=C{s}^2$

Substitute 0  for l(s)  into above equation.

  $C=\frac{I}{s^2}$

Now substitute   $\frac{1}{s^2}$for C  into above equation.

  $I\left(s\right)=I-I{\left(\frac{s}{a}\right)}^2$

Determine the magnetic field in the same direction as the case b.

$$\begin{gathered} B=\frac{{\mu }_{0}I}{2\pi s}\left[1-1+{\left(\frac{s}{a}\right)}^2\right] \\ =\frac{{\mu }_{0}I}{2\pi }\frac{s}{a^2} \end{gathered}$$  

Therefore, the value of magnetic field in the same direction as the case b is  .

 $\overrightarrow{B}=\frac{{\mu }_{0}I}{2\pi }\frac{s}{a^2}$