A scalar field $\displaystyle \phi $ is a function of x,z and t only. Vectors $\displaystyle {\bf{E}}$ and $\displaystyle {\bf{H}}$ are defined by:

$\displaystyle {\bf{E}} = \frac{1}{\varepsilon }\left( {\frac{{\partial \phi }}{{\partial {z^{}}}}{\bf{i}} - \frac{{\partial \phi }}{{\partial x}}{\bf{k}}} \right)$

$\displaystyle {\bf{H}} = - \frac{{\partial \phi }}{{\partial t}}{\bf{j}}$

The divergence of a vector field

$\displaystyle {\bf{F}} = {F_1}{\bf{i}} + {F_2}{\bf{j}} + {F_3}{\bf{k}}$

is a scalar given by:

$\displaystyle \nabla \cdot {\bf{F}} = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial t}} + \frac{{\partial {F_3}}}{{\partial z}}$

Show that

$\displaystyle \nabla \cdot {\bf{H}} = 0$

$\displaystyle \nabla \cdot {\bf{H}} = \left( {{\bf{i}}\frac{\partial }{{\partial x}} + {\bf{j}}\frac{\partial }{{\partial t}} + {\bf{k}}\frac{\partial }{{\partial z}}} \right) \cdot \left( { - \frac{{\partial \phi }}{{\partial {t^{}}}}{\bf{j}}} \right)$

$\displaystyle \frac{{\partial {F_2}}}{{\partial t}} = 0$

And show that $\displaystyle \nabla \cdot {\bf{E}} = 0$

$\displaystyle {\bf{E}} = \frac{1}{\varepsilon }\frac{{\partial \phi }}{{\partial {z^{}}}}{\bf{i}} - \frac{1}{\varepsilon }\frac{{\partial \phi }}{{\partial x}}{\bf{k}}$

$\displaystyle \nabla \cdot {\bf{E}} = \left( {{\bf{i}}\frac{\partial }{{\partial x}} + {\bf{j}}\frac{\partial }{{\partial z}} + {\bf{k}}\frac{\partial }{{\partial t}}} \right) \cdot \left( {\frac{1}{\varepsilon }\frac{{\partial \phi }}{{\partial {z^{}}}}{\bf{i}} - \frac{1}{\varepsilon }\frac{{\partial \phi }}{{\partial x}}{\bf{k}}} \right)$

So that gives

$\displaystyle \frac{{\partial {F_1}}}{{\partial x}} = \frac{1}{\varepsilon }\frac{{\partial \phi }}{{\partial {z^{}}}}$

$\displaystyle \frac{{\partial {F_3}}}{{\partial x}} = - \frac{1}{\varepsilon }\frac{{\partial \phi }}{{\partial x}}$

Which will cancel each other out

Given that

$\displaystyle \nabla \times {\bf{E}} = - \mu \frac{{\partial {\bf{H}}}}{{\partial t}}$

where u is a constant show that $\displaystyle \phi $ satisfies the partial differential equation

$\displaystyle \frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} = \mu \varepsilon \frac{{{\partial ^2}\phi }}{{\partial {t^2}}}$

How to go about this last part?

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