Divergence of ##\vec{x}/\vert\vec{x}\vert^3##

[PhysicsRock]

Homework Statement

Calculate $\displaystyle \frac{\partial}{\partial x_i} \frac{x_i}{\vert\vec{x}\vert^3}$

Relevant Equations

$\nabla \cdot \vec{F} = \sum_i \frac{\partial f_i}{\partial x_i}$

As you can see in the homework statement, I am asked to calculate what's effectively the divergence of the vector field $\vec{F} = \vec{x}/\vert\vec{x}\vert^3$ over $\mathbb{R}^3$. I have done that, the calculation itself isn't that difficult after all. However, I can't make sense of the result, which makes me wonder whether I've made a mistake. What I find is that $\nabla \cdot \vec{F} = 0$. I used GeoGebra to plot the field, and what I see is vectors "coming out" of the origin. Is it as simple as to say that, since you can't divide by 0, the origin isn't part of the field, and thus, there is no particular source? Or have I made a mistake in my calculations and the result isn't 0?

 

[BvU]

See this thread

Homework Statement: Calculate the following $\vec{\nabla}\cdot \vec{E}(\vec{r})$
Relevant Equations: none

Hi,

unfortunately, I am not sure if I have calculated the task correctly

View attachment 325180
The electric field of a point charge looks like this $\vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0}\frac{\vec{r}}{|\vec{r}|^3}$ I have now simply divided the electric field into its components i.e. #E_x , E-y, E_z#.

$$\vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0}\left( \begin{array}{rrr}
\frac{x}{(x^2+y^2+z^3)^{\frac{3}{2}}} \\
\frac{y}{(x^2+y^2+z^3)^{\frac{3}{2}}} \\
\frac{z}{(x^2+y^2+z^3)^{\frac{3}{2}}} \\
\end{array}\right)$$

Then I calculated the divergence

$$\vec{\nabla}\cdot \vec{E}(\vec{r})=\frac{Q}{4 \pi \epsilon_0} \Bigl( \frac{\partial}{\partial x}\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2} }} +\frac{\partial}{\partial y}\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2} }} +\frac{\partial}{\partial z}\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2} }} \Bigr)=\Bigl( \frac{-2x^2+y^2+z^2}{(x^2+y^2+z^2)^{\frac{5}{2} }} +\frac{x^2-2y^2+z^2}{(x^2+y^2+z^2)^{\frac{5}{2} }} +\frac{x^2+y^2-2z^2}{(x^2+y^2+z^2)^{\frac{5}{2} }} \Bigr)=0$$

With the result of 0 I am a bit confused, in the task is not mentioned the sign of the charge, but with a positive charge I would expect as a divergence a source, so the divergence would have to be positive and with a negative charge, a sink, so a negative divergence.

Have I somehow miscalculated, or do I have a thinking error in the physical interpretation of the result?

 

[fresh_42]

The vector field is indeed solenoidal, without divergence. All vector fields of type $\dfrac{\vec{c}\times \vec{r}}{\|\vec{r}\|^3}$ are.

Is it as simple as to say that, since you can't divide by 0, the origin isn't part of the field, and thus, there is no particular source? Or have I made a mistake in my calculations and the result isn't 0?

We always have a ball of some positive radius and vectors pointing outward of equal length. You could as well say that there is no way to lose or gain energy on a closed path through that field.

 

[PhysicsRock]

@fresh_42, @BvU thank you. Glad to hear I didn't screw up.

 

[FactChecker]

I am asked to calculate what's effectively the divergence of the vector field $\vec{F} = \vec{x}/\vert\vec{x}\vert^3$ over $\mathbb{R}^3$.

I think that the origin should be omitted.

Is it as simple as to say that, since you can't divide by 0, the origin isn't part of the field, and thus, there is no particular source?

IMO, you are correct. The initial statement of the problem seems a little careless by including the origin.
I would like to hear the opinion on this of others who have more expertise.