# Field from Potential (Homework 3-323)

3. Find the fields, the charge density and the current density corresponding to $V(\overrightarrow{r},t)=0$ and $A(\overrightarrow{r},t)=\frac{1}{4\pi\epsilon0}*\frac{e(t0-t)}{r^2}*\overrightarrow{r0}$ where $e$ and$t0$ are constants.

The potentials are:

$V(r ⃗,t)=0$ and $A ⃗(r ⃗,t)=1/(4πϵ_0 )*\frac{e(t_0-t)}{r^2} r_0 ⃗$ with $e$ and $t0$ constants

One has

$E ⃗=-∇V-(∂A ⃗)/∂t=0+[e/(4πϵ_0 r^2 )]r_0 ⃗$

$B=∇×A ⃗=e^2/(4πϵ_0 ) (∇×(r_0 ⃗/r^2 )$

Where

$∇×(r_0 ) ⃗/r^2 =[(∂/∂y) (z_0/(x^2+y^2+z^2 ))-(∂/∂z) (y_0/(x^2+y^2+z^2 )) ]i+⋯=$

$=2/r^4 [(y z_0-z y_0 )i+⋯]$

$∇×(r_0/r^2) =(2/r^4) (r ⃗ ×r_0 ⃗)$

So that

$B ⃗=e^2/(4πϵ_0 )*2/r^4 (r ⃗ ×r_0 ⃗)$

The charge density is

$1/ϵ_0 *ρ=-∇^2 V-∂/∂t (∇A ⃗ )$

$∇(r_0 ⃗/r^2 )=r_0 ⃗*∇(1/r^2 )=r_0 ⃗ (∂/∂x (1/(x^2+y^2+z^2 ))i+⋯)=$

$=r_0 ⃗ (-2x/r^4 i-…)=-(2r ⃗*r_0 ⃗)/r^4$

So that

$1/ϵ_0 *ρ=0-e/(4πϵ_0 ) ∇(r_0 ⃗/r^2 )=e/(4πϵ_0 )*(2r ⃗*r_0 ⃗)/r^4$

$ρ=(e/4π) [(2r ⃗*r_0 ⃗)/r^4]$

The current density is

$μ_0 J=(∇^2 A-1/c^2 (∂^2 A)/(∂t^2 ))-∇(∇A)=-1/c^2 (∂^2 A)/(∂t^2 )=0$