Is this vector calculus problem correct?

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?

Thanks for feedback.

Hey mark090480.

The first thing you should do is setup a matrix and calculate its determinant to calculate curl(E) and calculate the LHS accordingly. Based on this and the other constraint you will have two pieces of information which you can combine.

Hello Chiro,

Do you mean to solve that last part I should use the curl? The problem I have is that this is all the info I have, I cannot calculate the det as I have no real numbers to use. What about the other parts, are they looking good?

Thanks for any help.

You can do it symbolically just like you did the del(E). Just set up the matrix symbolically and then use algebra to calculate the determinant.