Derivative of the retarded vector potential

  • #1
Salmone
98
13
In a problem of an oscillating electric dipole, under appropriate conditions, one can find, for the potential vector calculated at the point ##\vec{r}##, the expression ##\vec{A}=\hat{k}\frac{\mu_0I_0d}{4\pi}\frac{cos(\omega(t-r/c))}{r}## where: ##\hat{k}## is the direction of the ##z-axis## where the dipole is oscillating, ##I_0## is the current (##I(t)=I_0cos(\omega t)##), ##d## is the distance between the charges of the dipole and ##r## is the distance between the origin of the system and the point where I want to calculate the potential vector. Let ##\vec{p}=\hat{k}qd=\frac{\hat{k}dI_0}{\omega}sin(\omega t)## be the dipole moment, it is possible to rewrite the potential vector as ##\vec{A}=\frac{\mu_0}{4\pi}\frac{\vec{\dot p(t-r/c)}}{r}## where ##\vec{\dot p}## is the derivative with respect to time.
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}##
 

Answers and Replies

  • #2
jasonRF
Science Advisor
Gold Member
1,505
572
When you took calculus, did you learn the chain rule?
 

Suggested for: Derivative of the retarded vector potential

Replies
1
Views
327
Replies
10
Views
438
Replies
10
Views
499
Replies
0
Views
179
Replies
2
Views
519
Replies
10
Views
474
Replies
2
Views
865
Replies
2
Views
666
  • Last Post
Replies
3
Views
341
Top