- #1

Salmone

- 101

- 13

If we use Lorentz Gauge we have ##-\epsilon_0\mu_0\frac{\partial V}{\partial t}=\vec{\nabla} \cdot \vec{A}##, since ##\vec{A}## is directed along the ##z-axis## we have ##div \vec{A}=\frac{\partial A}{\partial z}## so ##\frac{\partial V}{\partial t}=-\frac{1}{\epsilon_0 \mu_0}\frac{\partial A}{\partial z}##.

**The question: how do you do this last derivative?**

The result is: ##\frac{\partial V}{\partial t}=\frac{1}{4\pi\epsilon_0} \left( \frac{- \ddot p(t-r/c)}{cr}+\frac{\dot p(t-r/c)}{r^2} \right)\frac{z}{r}##