My teacher told me to solve some physics problem where I need to find $\rho(r)$ by using: $$\frac{\rho(r).dV}{\varepsilon_0}=\vec E(r+dr)S(r+dr)-\vec E(r)S(r)$$ where $dV$ is a small hollow spherical volume, i.e. $\frac d{dr}\left({\frac43\pi r^3}\right)=4\pi r^2dr$ and $\vec E(r)$ is a known vector (like $\vec E(r)=ar^4\hat r$ and a is a constant, after solving which we get $\rho=6ar^3\varepsilon_0$), and $S(r)$ is the spherical surface area $4\pi r^2$.

I found another convinient formula in a good book: $$\vec\nabla.\vec E=\frac{\rho}{\varepsilon_0}$$

I know that: $$\vec\nabla=\sum\frac{\partial}{\partial x}\hat x$$ and in spherical coordinates: $$\vec\nabla=\frac{\partial}{\partial r}\hat r+\frac{\partial}{r\partial \theta}\hat \theta+\frac{\partial}{r\sin\theta\partial \phi}\hat \phi$$ So I tried: $$\vec\nabla.\vec E=\left(\frac{\partial}{\partial r}\hat r+\frac{\partial}{r\partial \theta}\hat \theta+\frac{\partial}{r\sin\theta\partial \phi}\hat \phi\right).(ar^4\hat r)=\frac{\partial}{\partial r}ar^4=4ar^3\ne6ar^3$$

After checking back I found: $$\vec\nabla.\vec E=\frac{\partial(r^2E_r)}{r^2\partial r}+...=6ar^3$$

How did $r^2$ come there in?


The divergence should be calculated taking into account that the unit vector themselves are functions of the coordinates!

  • 1
    $\begingroup$ This would be so much better if you didn't have an image full of math, and instead typed out a summary. $\endgroup$
    – David Z
    Dec 11 '14 at 8:49
  • $\begingroup$ @DavidZ I would, when I have a spare time, but I'm very sorry I presently don't have $\endgroup$
    – RE60K
    Dec 11 '14 at 17:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.