- #1

netheril96

- 194

- 0

\nabla \cdot \frac{{\vec e_r }}{{r^2 }} = 4\pi \delta (\vec r)

\]

[/tex]

This can be seen from[tex]\[

\nabla \cdot \frac{{\vec e_r }}{{r^2 }} = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}(r^2 \cdot \frac{1}{{r^2 }}) = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}(1) = 0(r \ne 0)

\]

[/tex]

And from Gauss' Theorem[tex]\[

\int_V {(\nabla \cdot \frac{{\vec e_r }}{{r^2 }})dV = \oint_S {\frac{{\vec e_r }}{{r^2 }} \cdot d\vec S} } = 4\pi

\]

[/tex]

But if I want to directly using the formula of divergence in spherical coordinates,I can only get[tex]\[

\nabla \cdot \frac{{\vec e_r }}{{r^2 }} = \frac{1}{{r^2 }}\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})

\]

[/tex]

And integrating this over a volume cannot give me the result of 4π[tex]\[

\int_V {(\nabla \cdot \frac{{\vec e_r }}{{r^2 }})dV = } \int_0^\pi {\sin \theta d\theta \int_0^{2\pi } {d\phi \int_0^R {\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})} } } dr = 4\pi \int_0^R {\frac{\partial }{{\partial r}}(\frac{{r^2 }}{{r^2 }})} dr

\]

[/tex]

(Here V is a sphere with radius of R)

So how can I connect it with Dirac Delta?

By the way,I post this here because this problem arises in the electrostatic field of a point charge and I found nothing about such thing in any book concerning δ(x).