Question

b) Figure Q1(b) describes an infinite sheet of electric surface current density (L=a,JoA /m) directed along the x-axis, and placed on the z=0 plane between free space for the region z < 0 and a dielectric medium for z > 0. It is known from the solution of 'a) above that the electric field intensity excited by this distribution of current is described by E= provided that k,-ω μ,e°, k=0)Ju,e,e'. By imposing boundary conditions as may apply, determine the constants E and E2 in terms of Jo and the constitutive parameters. What is the expression for the corresponding magnetic field? Figure Question 1(b)

Answer 1

DATA:

$\\ \vec{J}_{s}=\left (J_{o}\,\,\widehat{a_{x}} \right )\,A/m \\ \\ \vec{E}=\left\{\begin{matrix} E_{1}\,e^{j\,k_{o}\,z}\,\widehat{a}_{x}\,\,;\,\,z< 0 & \\ E_{2}\,e^{j\,k\,z}\,\widehat{a}_{x}\,\,\,;\,\,z> 0 & \end{matrix}\right. \\ \\ k_{o}=\omega\sqrt{\mu_{o}\varepsilon_{o}} \\ k=\omega\sqrt{\mu_{o}\varepsilon_{o}\varepsilon_{r}}$

Solution

According to the following equation:

$\left ( \vec{E}_{2}-\vec{E}_{1} \right )\times \widehat{n}=0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

Where:

$\vec{E}_{1}=\left (E_{1}\,e^{jk_{o}z} \right )\,\,\widehat{a}_{x}\,\,\,\,\,\,for \,\,\,\,z< 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$

and

$\vec{E}_{2}=\left (E_{2}\,e^{j k z} \right )\,\,\widehat{a}_{x}\,\,\,\,\,\,for \,\,\,\,z> 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$

Inserting (2) and (3) into equation (1) we obtain:

Similarly, for the magnetic field we have:

$\widehat{a}_{z}\times\left ( \vec{H}_{2}-\vec{H}_{1} \right )=\vec{J}_{s}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (5)$

Where;

$\vec{H}_{1}=\left (-\frac{E_{1}}{\eta_{o}}e^{jk_{o}z} \right )\,\,\,\widehat{a}_{y}\,\,\,\,for\,\,\,z< 0\,\,\,\,\: \: \: \: \: \: \: \: \: \: \: \: (6)$

and

$\vec{H}_{2}=\left (\frac{E_{2}}{\eta}e^{jkz} \right )\,\,\,\widehat{a}_{y}\,\,\,\,for\,\,\,z> 0\,\,\,\,\: \: \: \: \: \: \: \: \: \: \: \: (7)$

With   $\eta_{o}$ and   $\eta$ constant values as follows:

$\eta_{o}=\sqrt{\frac{\mu_{o}}{\varepsilon_{o}}}\,\,\,\,\,\,\,\,\,\,\, \, \, \, \, \, \, \, \, \, (8)$

and

$\eta=\sqrt{\frac{\mu_{o}}{\varepsilon_{o}\varepsilon_{r}}}\,\,\,\,\,\,\,\,\,\,\, \, \, \, \, \, \, \, \, \, (9)$

Replacing (6) and (7) into equation (5) we have:

$\widehat{a}_{z}\times\left [ \frac{E_{2}}{\eta}e^{jkz}-\left ( -\frac{E_{1}}{\eta_{o}} \right )e^{jk_{o}z} \right ]\widehat{a}_{y}=\vec{J}_{s}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (10)$

But:

$\vec{J}_{s}=J_{o}\,\,\widehat{a}_{x}\: \: \: \: \: \: \: \: \, \, \, \, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (11)$

Inserting (11) into (10):

$\widehat{a}_{z}\times\left [ \frac{E_{2}}{\eta}e^{jkz}-\left ( -\frac{E_{1}}{\eta_{o}} \right )e^{jk_{o}z} \right ]\widehat{a}_{y}=J_{o}\,\,\widehat{a}_{x}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (11)$

Solving we obtain:

$-\left ( \frac{E_{2}}{\eta}\,+\,\frac{E_{1}}{\eta_{o}} \right )=J_{o}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (12)$

Inserting (4) into (12) we have:

$-\left ( \frac{E_{1}}{\eta}\,+\,\frac{E_{1}}{\eta_{o}} \right )=J_{o}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (13)$

Isolating $E_{1}$ we obtain:

$E_{1}=-\frac{J_{o}\eta_{o}\eta}{\eta_{o}+\eta}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (14)$

Therefore:

${\color{Blue} E_{1}=E_{2}=-\frac{J_{o}\eta_{o}\eta}{\eta_{o}+\eta}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (15)$

Now, the magnetic field can be written as:

$\vec{H}=\left\{\begin{matrix} \left ( \frac{J_{o}\eta}{\eta_{o}+\eta} \right )e^{jk_{o}z}\,\,\widehat{a}_{y}\: \: \: \: z< 0 & \\ \\ -\left ( \frac{J_{o}\eta_{o}}{\eta_{o}+\eta} \right )e^{jkz}\,\,\widehat{a}_{y}\, \, \, \,\, z> 0 & \end{matrix}\right.$