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I was going through MIT opencourseware https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-013-electromagnetics-and-applications-spring-2009/readings/MIT6_013S09_chap07.pdf and I didnt understand exactly why in equation 7.1.37 and 7.1.38 in page 192 the divergence of transverse electric field is taken as zero ? The only reason this would be true is if the region is source free but we clearly see that this field does setup a surface charge density on the surface of the transmission line so intuitively and mathematically why would the divergence of Et be zero?

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For example, according to 7.1.37 $$\text{div}\textbf{H}=\text{div}_T\textbf{H}_T+\frac{\partial}{\partial z}\hat {\textbf{z}} \cdot \textbf{H}_T $$ but you know that there are no magnetic space charges $\text{div}\textbf{H}=0$ and by definition $\hat {\textbf{z}} \cdot \textbf{H}_T$, therefore you must also have $$\text{div}_T\textbf{H}_T=0$$; same consideration for the $\textbf{E}$ field.

The induced surface charges or currents play no role in the spatial divergence of the fields but they are important to define the boundary conditions: the E-field is perpendicular and $\propto \sigma$ and H-field is tangential with the ideal metal surface and $\propto \K$.

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Here's my understanding. This is an excellent question, which i think should attract a lot of discussion.

The question is about TEM wave between two conductors, which are not necessarily planar, but having a cross-section that is independent of z.

//In planar (parallel plate tx line) case, we have followed these steps:

-set \$\mathbf{E}\$ and \$\mathbf{H}\$, such that the general boundary condition requirement (\$\mathbf{E}\$ perp, \$\mathbf{H}\$ parallel, to metal) is satisfied. Our assumption that E and H field are TEM (\$\mathbf{E}=\mathbf{E}_T, \mathbf{H}=\mathbf{H}_T\$), as well as, uniform plane wave (\$\mathbf{E}=\mathbf{E}_T(z,t)=\hat{x}E_0cos(wt-kz)\$ and \$\mathbf{H}=\mathbf{H}_T(z,t)\$, in eq.7.1.1) satisfy these constraints.

-find \$\mathbf{E}\$ and \$\mathbf{H}\$ wave equation, as they travel in the medium between the planes, which is indeed a source free region, so \$J=0=Q\$, and \$\mathbf{\nabla}.\mathbf{E}=0\$. Verify: \$E=\hat{x}E_0cos(wt-kz) \implies \mathbf{\nabla}.\mathbf{E}=\frac{\partial E_x}{\partial x}+0+0=0\$. For TEM, \$\mathbf{\nabla}.\mathbf{E}=\mathbf{\nabla _T}.\mathbf{E_T}\$, so \$\mathbf{\nabla _T}.\mathbf{E_T}=0\$ follows.

-The \$\mathbf{E}\$ and \$\mathbf{H}\$ so obtained, must be supported by surface charges and currents at the metal surface, so we use the (integral form of) \$\mathbf{\nabla}.\mathbf{E}=\pm Q/\epsilon_0\$ at \$x=0\$, and \$x=d\$, to find \$Q\$ in terms of \$\mathbf{E}\$ [then find \$V\$ in terms of \$\mathbf{E}\$, thus find Capacitance \$C\$].

//Similar logic and steps for non-planar tx line case:

-set \$\mathbf{E}\$ and \$\mathbf{H}\$, such that the general boundary condition requirement (\$\mathbf{E}\$ perp, \$\mathbf{H}\$ parallel, to metal) is satisfied. But now the two surfaces are arbitrary, so we can no longer impose plane wave form, but only TEM form: \$\mathbf{E}=\mathbf{E}_T(x,y,z,t)\$ and \$\mathbf{H}=\mathbf{H}_T(x,y,z,t)\$.

-find \$\mathbf{E}\$ and \$\mathbf{H}\$ wave equation, as they travel in the medium between the planes, which is indeed a source free region, so \$J=0=Q\$, and \$\mathbf{\nabla}.\mathbf{E}=0\$. For TEM, \$\mathbf{\nabla}.\mathbf{E}=\mathbf{\nabla _T}.\mathbf{E_T}\$, so \$\mathbf{\nabla _T}.\mathbf{E_T}=0\$ follows.

-The \$\mathbf{E}\$ and \$\mathbf{H}\$ so obtained, must be supported by surface charges and currents at the metal surface, so we use the (integral form of) \$\mathbf{\nabla}.\mathbf{E}=\pm Q/\epsilon_0\$ on a Gaussian surface covering one of the conductors, to find \$Q\$ in terms of \$\mathbf{E}\$ [in eq 7.1.39, then find \$V\$ in terms of \$\mathbf{E}\$, thus find Capacitance \$C\$].

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