Let's solve for the electric and magnetic fields produced by the current sheet \[ \mathbf{J}_S(x, y) = \hat{y} J_0 e^{-jk_x x} e^{-jk_y y} \] located in the \(z = 0\) plane in free space. The fields are assumed to have the forms: - For \(z > 0\): \[ \mathbf{E}_1 = (\hat{x} E_{1x} + \hat{y} E_{1y} + \hat{z} E_{1z}) e^{-jk_x x - jk_y y - jk_z z} \] - For \(z < 0\): \[ \mathbf{E}_2 = (\hat{x} E_{2x} + \hat{y} E_{2y} + \hat{z} E_{2z}) e^{-jk_x x - jk_y y + jk_z z} \] We are to determine the coefficients using Maxwell’s equations and boundary conditions. --- ## 1. **Plane Wave Structure and Dispersion Relation** Since the fields are plane waves, \[ k^2 = k_x^2 + k_y^2 + k_z^2 \] where \(k = \omega \sqrt{\mu_0 \epsilon_0}\). --- ## 2. **Express \(\mathbf{H}_1\) and \(\mathbf{H}_2\) Using Maxwell's Equations** From Maxwell’s curl equation in free space: \[ \nabla \times \mathbf{E} = -j\omega\mu_0 \mathbf{H} \] For a plane wave of the form \(e^{-j\mathbf{k}\cdot\mathbf{r}}\): \[ \nabla \to -j\mathbf{k} \] so \[ -j\mathbf{k} \times \mathbf{E} = -j\omega\mu_0 \mathbf{H} \implies \mathbf{H} = \frac{1}{\omega\mu_0} \mathbf{k} \times \mathbf{E} \] Let’s write out the vector \(\mathbf{k}\): - For \(z > 0\): \(\mathbf{k}_1 = k_x \hat{x} + k_y \hat{y} + k_z \hat{z}\) - For \(z < 0\): \(\mathbf{k}_2 = k_x \hat{x} + k_y \hat{y} - k_z \hat{z}\) So - \(\mathbf{H}_1 = \frac{1}{\omega\mu_0} \mathbf{k}_1 \times \mathbf{E}_1\) - \(\mathbf{H}_2 = \frac{1}{\omega\mu_0} \mathbf{k}_2 \times \mathbf{E}_2\) --- ## 3. **Boundary Conditions at \(z=0\)** **a) Tangential \(\mathbf{E}\) continuous:** \[ \boxed{ \mathbf{\hat{z}} \times \left[ \mathbf{E}_1 - \mathbf{E}_2 \right]_{z=0} = 0 } \] This means the \(x\) and \(y\) components of \(\mathbf{E}\) are continuous across the sheet. **b) Discontinuity in tangential \(\mathbf{H}\) due to surface current:** \[ \boxed{ \mathbf{\hat{z}} \times \left[ \mathbf{H}_1 - \mathbf{H}_2 \right]_{z=0} = \mathbf{J}_S } \] **c) Discontinuity in normal \(\mathbf{E}\) due to surface charge:** \[ \mathbf{\hat{z}} \cdot [\mathbf{E}_1 - \mathbf{E}_2]_{z=0} = \frac{\rho_s}{\epsilon_0} \] where \(\rho_s\) is a surface charge density. --- ## 4. **Write Out Field Components** Let \[ \mathbf{E}_1 = (E_{1x}\hat{x} + E_{1y}\hat{y} + E_{1z}\hat{z}) e^{-jk_x x - jk_y y - jk_z z} \] \[ \mathbf{E}_2 = (E_{2x}\hat{x} + E_{2y}\hat{y} + E_{2z}\hat{z}) e^{-jk_x x - jk_y y + jk_z z} \] Similarly, for \(\mathbf{H}\): For \(z>0\): \[ \mathbf{H}_1 = \frac{1}{\omega\mu_0} \left[ \begin{array}{l} (k_y E_{1z} - k_z E_{1y})\hat{x} \\ (k_z E_{1x} - k_x E_{1z})\hat{y} \\ (k_x E_{1y} - k_y E_{1x})\hat{z} \end{array} \right] e^{-jk_x x-jk_y y - jk_z z} \] For \(z<0\): \[ \mathbf{H}_2 = \frac{1}{\omega\mu_0} \left[ \begin{array}{l} (k_y E_{2z} + k_z E_{2y})\hat{x} \\ (-k_z E_{2x} - k_x E_{2z})\hat{y} \\ (k_x E_{2y} - k_y E_{2x})\hat{z} \end{array} \right] e^{-jk_x x-jk_y y + jk_z z} \] Note the sign changes for \(k_z\) in \(\mathbf{k}_2\). --- ## 5. **Apply Boundary Conditions** ### a) Tangential \(\mathbf{E}\) Continuous At \(z=0\): \[ E_{1x} = E_{2x}, \qquad E_{1y} = E_{2y} \] Let’s call them: \[ E_x \equiv E_{1x} = E_{2x},\qquad E_y \equiv E_{1y} = E_{2y} \] ### b) Tangential \(\mathbf{H}\) Discontinuity \[ \mathbf{\hat{z}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_S \] Calculate \(\mathbf{\hat{z}} \times \mathbf{H}\): \[ \mathbf{\hat{z}} \times \mathbf{H} = (\hat{x} H_y - \hat{y} H_x) \] So at \(z=0\): \[ \mathbf{\hat{z}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \hat{x} (H_{1y} - H_{2y}) - \hat{y} (H_{1x} - H_{2x}) \] But \(\mathbf{J}_S = \hat{y} J_0 e^{-jk_x x - jk_y y}\), so only the \(\hat{y}\) component is nonzero: \[ \hat{x}: \quad H_{1y} - H_{2y} = 0 \] \[ -\hat{y}: \quad -(H_{1x} - H_{2x}) = J_0 e^{-jk_x x - jk_y y} \implies H_{1x} - H_{2x} = -J_0 e^{-jk_x x - jk_y y} \] --- ## 6. **Write Out the Relations** ### a) \(H_{1y} - H_{2y} = 0\): Using formulas above: \[ H_{1y} = \frac{1}{\omega\mu_0}(k_z E_x - k_x E_{1z}) \] \[ H_{2y} = \frac{1}{\omega\mu_0}(-k_z E_x - k_x E_{2z}) \] So: \[ \frac{1}{\omega\mu_0}(k_z E_x - k_x E_{1z} + k_z E_x + k_x E_{2z}) = 0 \implies \frac{1}{\omega\mu_0}(2 k_z E_x - k_x(E_{1z} - E_{2z})) = 0 \] \[ 2 k_z E_x - k_x (E_{1z} - E_{2z}) = 0 \] \[ E_{1z} - E_{2z} = \frac{2 k_z}{k_x} E_x \] ### b) \(H_{1x} - H_{2x} = -J_0 e^{-jk_x x - jk_y y}\): \[ H_{1x} = \frac{1}{\omega\mu_0}(k_y E_{1z} - k_z E_y) \] \[ H_{2x} = \frac{1}{\omega\mu_0}(k_y E_{2z} + k_z E_y) \] Then, \[ H_{1x} - H_{2x} = \frac{1}{\omega\mu_0}[k_y(E_{1z} - E_{2z}) - k_z(E_y + E_y)] = \frac{1}{\omega\mu_0}[k_y(E_{1z} - E_{2z}) - 2k_z E_y] \] So set equal to \(-J_0 e^{-jk_x x - jk_y y}\): \[ \frac{1}{\omega\mu_0}[k_y(E_{1z} - E_{2z}) - 2k_z E_y] = -J_0 \] Plug in \(E_{1z} - E_{2z} = \frac{2 k_z}{k_x} E_x\): \[ \frac{1}{\omega\mu_0}\left[k_y \frac{2 k_z}{k_x} E_x - 2k_z E_y\right] = -J_0 \] \[ \frac{2k_z}{\omega\mu_0}\left(\frac{k_y}{k_x} E_x - E_y\right) = -J_0 \] --- ## 7. **Divergence Condition** From Maxwell's equations: \(\nabla \cdot \mathbf{E} = 0\) in free space. For the assumed plane wave form, \(\mathbf{k} \cdot \mathbf{E} = 0\): So for both half-spaces: \[ k_x E_x + k_y E_y \pm k_z E_{1z,2z} = 0 \] At \(z=0\): \[ k_x E_x + k_y E_y + k_z E_{1z} = 0 \] \[ k_x E_x + k_y E_y - k_z E_{2z} = 0 \] --- ## 8. **Solve for Coefficients** Let’s solve for \(E_x, E_y, E_{1z}, E_{2z}\): Let’s define everything in terms of \(E_y\): From above, \(E_{1z} - E_{2z} = \frac{2k_z}{k_x}E_x\). From divergence in \(z>0\): \[ k_x E_x + k_y E_y + k_z E_{1z} = 0 \implies E_{1z} = -\frac{k_x}{k_z} E_x - \frac{k_y}{k_z} E_y \] Similarly in \(z<0\): \[ k_x E_x + k_y E_y - k_z E_{2z} = 0 \implies E_{2z} = \frac{k_x}{k_z} E_x + \frac{k_y}{k_z} E_y \] Thus, \[ E_{1z} - E_{2z} = -\frac{2k_x}{k_z} E_x - \frac{2k_y}{k_z} E_y \] But from before: \[ E_{1z} - E_{2z} = \frac{2k_z}{k_x} E_x \] So, \[ \frac{2k_z}{k_x} E_x = -\frac{2k_x}{k_z} E_x - \frac{2k_y}{k_z} E_y \] Solve for \(E_x\): \[ \frac{k_z}{k_x} E_x + \frac{k_x}{k_z} E_x + \frac{k_y}{k_z} E_y = 0 \] Multiply by \(k_z\): \[ \frac{k_z^2}{k_x} E_x + k_x E_x + k_y E_y = 0 \] \[ \left(\frac{k_z^2}{k_x} + k_x\right) E_x + k_y E_y = 0 \] But \(k^2 = k_x^2 + k_y^2 + k_z^2\), so \(k_z^2 = k^2 - k_x^2 - k_y^2\): \[ \frac{k^2 - k_x^2 - k_y^2}{k_x} E_x + k_x E_x + k_y E_y = 0 \] \[ \frac{k^2 - k_x^2 - k_y^2 +

# Step-by-Step Solution for Fields from a Surface Current Sheet ## 1. **Problem Summary** - Surface current density: \[ \mathbf{J}_S(x, y) = \hat{y} J_0 e^{-jk_x x} e^{-jk_y y} \] - Location: \(z=0\) plane in free space - Fields: \[ \mathbf{E}_1, \quad z>0; \quad \mathbf{E}_2, \quad z<0 \] with plane wave dependence: \[ \mathbf{E}_1 = \mathbf{E}_1(z) e^{-jk_x x - jk_y y}, \quad \mathbf{E}_2 = \mathbf{E}_2(z) e^{-jk_x x - jk_y y} \] - Goal: Find \(\mathbf{E}\) and \(\mathbf{H}\) fields satisfying Maxwell’s equations and boundary conditions. --- ## 2. **Assumed Field Forms** - For \(z>0\): \[ \mathbf{E}_1 = (\hat{x} E_{1x} + \hat{y} E_{1y} + \hat{z} E_{1z}) e^{-jk_x x - jk_y y - jk_z z} \] - For \(z<0\): \[ \mathbf{E}_2 = (\hat{x} E_{2x} + \hat{y} E_{2y} + \hat{z} E_{2z}) e^{-jk_x x - jk_y y + jk_z z} \] where \(k_z\) is the z-component of the wavevector. --- ## 3. **Maxwell’s Equations in Frequency Domain** - In free space: \[ \nabla \times \mathbf{E} = -j \omega \mu_0 \mathbf{H} \] - Plane wave solutions: \[ \mathbf{H} = \frac{1}{\omega \mu_0} \mathbf{k} \times \mathbf{E} \] - Wavevector: \[ \mathbf{k}_1 = (k_x, k_y, k_z), \quad \mathbf{k}_2 = (k_x, k_y, -k_z) \] --- ## 4. **Boundary Conditions at \(z=0\)** ### a) **Tangential \(\mathbf{E}\) Continuous:** \[ E_{1x} = E_{2x} = E_x,\quad E_{1y} = E_{2y} = E_y \] ### b) **Discontinuity in \(\mathbf{H}_t\):** \[ \mathbf{\hat{z}} \times (\mathbf{H}_1 - \mathbf{H}_2) = \mathbf{J}_S \] Expanding: \[ (H_{1y} - H_{2y}) \hat{x} - (H_{1x} - H_{2x}) \hat{y} = \hat{y} J_0 e^{-jk_x x - jk_y y} \] which implies: \[ H_{1x} - H_{2x} = -J_0 e^{-jk_x x - jk_y y} \] \[ H_{1y} - H_{2y} = 0 \] --- ## 5. **Expressing \(\mathbf{H}\) in Terms of \(\mathbf{E}\)** Using \(\nabla \times \mathbf{E} = -j \omega \mu_0 \mathbf{H}\), for plane waves: \[ \mathbf{H} = \frac{1}{\omega \mu_0} \mathbf{k} \times \mathbf{E} \] - For \(z>0\): \[ \mathbf{H}_1 = \frac{1}{\omega \mu_0} \mathbf{k}_1 \times \mathbf{E}_1 \] \[ \Rightarrow \begin{cases} H_{1x} = \frac{1}{\omega \mu_0}(k_y E_{1z} - k_z E_{1y}) \\ H_{1y} = \frac{1}{\omega \mu_0}(k_z E_{1x} - k_x E_{1z}) \\ H_{1z} = \frac{1}{\omega \mu_0}(k_x E_{1y} - k_y E_{1x}) \end{cases} \] - Similarly for \(z<0\): \[ \mathbf{H}_2 = \frac{1}{\omega \mu_0} \mathbf{k}_2 \times \mathbf{E}_2 \] with \(k_z \to -k_z\). --- ## 6. **Applying Boundary Conditions** ### a) **Continuity of \(E_x, E_y\):** \[ E_{1x} = E_{2x} = E_x, \quad E_{1y} = E_{2y} = E_y \] ### b) **Discontinuity in \(H_x\):** \[ H_{1x} - H_{2x} = -J_0 e^{-jk_x x - jk_y y} \] Express \(H_{1x}\) and \(H_{2x}\): \[ H_{1x} = \frac{1}{\omega \mu_0}(k_y E_{1z} - k_z E_{1y}) \] \[ H_{2x} = \frac{1}{\omega \mu_0}(k_y E_{2z} + k_z E_{2y}) \] Since \(E_{1y} = E_{2y} = E_y\), the difference: \[ H_{1x} - H_{2x} = \frac{1}{\omega \mu_0} [k_y (E_{1z} - E_{2z}) - k_z (E_{1y} + E_{2y})] \] but \(E_{1y} = E_{2y} = E_y\), so: \[ H_{1x} - H_{2x} = \frac{1}{\omega \mu_0} [k_y (E_{1z} - E_{2z}) - 2 k_z E_y] \] Set equal to \(-J_0 e^{-jk_x x - jk_y y}\): \[ \frac{1}{\omega \mu_0} [k_y (E_{1z} - E_{2z}) - 2 k_z E_y] = -J_0 \] \[ k_y (E_{1z} - E_{2z}) - 2 k_z E_y = -\omega \mu_0 J_0 \] --- ## 7. **Relate \(E_{z}\) Components** From divergence of \(\mathbf{E}\): \[ k_x E_x + k_y E_y + k_z E_{1z} = 0 \Rightarrow E_{1z} = -\frac{k_x}{k_z} E_x - \frac{k_y}{k_z} E_y \] \[ E_{2z} = -\frac{k_x}{k_z} E_x - \frac{k_y}{k_z} E_y \] Thus: \[ E_{1z} - E_{2z} = 0 \] which simplifies the previous relation: \[ 0 - 2 k_z E_y = - \omega \mu_0 J_0 \] \[ E_y = \frac{\omega \mu_0 J_0}{2 k_z} \] --- ## 8. **Determine \(E_x\) and \(E_z\)** Using divergence in \(z>0\): \[ E_{1z} = -\frac{k_x}{k_z} E_x - \frac{k_y}{k_z} E_y \] Similarly, for \(z<0\): \[ E_{2z} = -\frac{k_x}{k_z} E_x - \frac{k_y}{k_z} E_y \] --- ## 9. **Final Expressions for Fields** ### Electric field components: \[ \boxed{ \begin{aligned} E_x &= \text{arbitrary (can choose based on polarization)} \\ E_y &= \frac{\omega \mu_0 J_0}{2 k_z} \\ E_z &= -\frac{k_x}{k_z} E_x - \frac{k_y}{k_z} E_y \end{aligned} } \] ### Magnetic field components: \[ \mathbf{H}_1 = \frac{1}{\omega \mu_0} \mathbf{k}_1 \times \mathbf{E}_1 \] \[ \mathbf{H}_2 = \frac{1}{\omega \mu_0} \mathbf{k}_2 \times \mathbf{E}_2 \] where \(\mathbf{k}_1 = (k_x, k_y, k_z)\), \(\mathbf{k}_2 = (k_x, k_y, -k_z)\). --- ## 10. **Summary** - The tangential electric fields are continuous and characterized by free parameters \(E_x\) (polarization choice) and known \(E_y\). - The surface current induces a discontinuity in \(H_x\), which determines \(E_y\). - The resulting fields are plane waves with the specified wavevector components, satisfying boundary conditions and Maxwell’s equations. --- **Note:** There is a nonzero surface charge density implied by the discontinuity in the normal component of \(\mathbf{E}\), consistent with the presence of a surface current. --- ## **Final note:** The explicit field expressions depend on the chosen polarization \(E_x\). Typically, one sets a polarization basis, then computes all components accordingly.