- #1

TheGreatDeadOne

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- Homework Statement
- Let a sphere of radius ##r_0## be centered at the origin, and ##\vec{r'}## the position vector of a point p' within the sphere or under its surface S. Let the position vector ##\vec{r}## be an arbitrary fixed point P. With this information, solve the integral, and analyze the cases for ##r\geq r_0## and ##r\leq r_0##

- Relevant Equations
- $$I=\oint_{s} \frac{1}{|\vec{r}-\vec{r'}|}da'$$

Solving the integral is the easiest part. Using spherical coordinates:

$$ \oint_{s} \frac{1}{|\vec{r}-\vec{r'}|}da' = \int_{0}^{\pi}\int_{0}^{2\pi} \frac{1}{|\vec{r}-\vec{r'}|}r_{0}^2 \hat r \sin{\theta}d\theta d\phi$$

then:

$$I = \dfrac{1}{|\vec{r}-\vec{r'}|}r_{0}^2(1+1)(2\pi)\hat r=\dfrac{4\pi r_{0}^2}{|\vec{r}-\vec{r'}|}\hat r $$

But what I couldn't understand was the final answer which is:

##\frac{4\pi r^{2}_{0}}{r} \quad\text{for} \quad r\geq r_0 \quad and \quad 4\pi r^{2}_{0}\quad for \quad r\leq r_0 ##

What happened to ##\frac{1}{|\vec{r}-\vec{r'}|} \hat r## for both ## r\geq r_0## and ## r\leq r_0 ## ? I know that for r -> 0 the integral explodes -Dirac enters here? -, but I don't know how this can help to reach the final result and I was very confused by the analysis for ## r\geq r_0## and ## r\leq r_0 ## part.

$$ \oint_{s} \frac{1}{|\vec{r}-\vec{r'}|}da' = \int_{0}^{\pi}\int_{0}^{2\pi} \frac{1}{|\vec{r}-\vec{r'}|}r_{0}^2 \hat r \sin{\theta}d\theta d\phi$$

then:

$$I = \dfrac{1}{|\vec{r}-\vec{r'}|}r_{0}^2(1+1)(2\pi)\hat r=\dfrac{4\pi r_{0}^2}{|\vec{r}-\vec{r'}|}\hat r $$

But what I couldn't understand was the final answer which is:

##\frac{4\pi r^{2}_{0}}{r} \quad\text{for} \quad r\geq r_0 \quad and \quad 4\pi r^{2}_{0}\quad for \quad r\leq r_0 ##

What happened to ##\frac{1}{|\vec{r}-\vec{r'}|} \hat r## for both ## r\geq r_0## and ## r\leq r_0 ## ? I know that for r -> 0 the integral explodes -Dirac enters here? -, but I don't know how this can help to reach the final result and I was very confused by the analysis for ## r\geq r_0## and ## r\leq r_0 ## part.

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