Question goes as

Show that the solutions to the Maxwell's equations $$ \nabla \times \vec H = \frac 1 c \frac{\partial \vec E}{\partial t}+\frac {4\pi} c \vec J, \hspace{ 2 mm} \nabla \times \vec E = -\frac 1 c \frac{\partial \vec H}{\partial t}, \hspace{ 2 mm} \nabla \cdot \vec H = 0, \hspace{ 2 mm} \nabla \cdot \vec E = 4 \pi \rho $$ are given by $$ \vec E = -\nabla \phi - \frac 1 c \frac{\partial \vec A}{\partial t}, \hspace{2mm} \vec H = \nabla \times \vec A,$$ where $ \phi$ and $\vec A$ are scalar potential and vector potential respectively which satisfy the equations $$ 1) \nabla \cdot \vec A + \frac 1c \frac{\partial \phi}{\partial t} = 0 $$ $$2) \nabla^2 \phi - \frac 1 {c^2} \frac{\partial ^2 \phi}{\partial t^2} = -4\pi \rho $$ $$3) \nabla^2 \vec A - \frac{1}{c^2}\frac{\partial ^2 \vec A}{\partial t^2}= -\frac {4\pi} c \vec J$$

Hint requested thank you!!
Also what kind of regions is described by $$ \nabla \cdot \vec E = 4 \pi \rho?$$ It is uniformly charged region?


1 Answer 1


Unless I'm missing something, all you have to do is substitute the solution you're given into Maxwell's equations and confirm the solution satisfies the equations.

For example you're given the solution $\vec H = \nabla \times \vec A$. Substitute this into $\nabla \cdot \vec H = 0$ and you get:

$$ \nabla \cdot (\nabla \times \vec A) = 0 $$

and this is true because div of curl is always zero.

Re comment: we make a point of not doing homework for you because your professor wouldn't be amused. You need to start working away at the equations and post here if you get stuck, giving the details of what's blocking you.

I showed the solution you're given satisfies $\nabla \cdot \vec H = 0$ because that's pretty trivial. I suggest you next tackle $\nabla \cdot \vec E = 4 \pi \rho$. Note that you're given:

$$ \nabla^2 \phi - \frac 1 {c^2} \frac{\partial ^2 \phi}{\partial t^2} = -4\pi \rho $$

so you just have to show that:

$$ -\nabla \cdot \vec E = \nabla^2 \phi - \frac 1 {c^2} \frac{\partial ^2 \phi}{\partial t^2} $$

You can do this using the information you're given in the question. A clue is that div of grad is del squared.

  • $\begingroup$ yeah that's the only part i know!! ... how to get that $ \vec E$ part $\endgroup$
    – S L
    Jul 1, 2012 at 12:14

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