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Problem:
For the vector function \vec{F}(\vec{r})=\frac{r\hat{r}}{(r^2+{\epsilon}^2)^{3/2}}
a. Calculate the divergence of ##\vec{F}(\vec{r})##, and sketch a plot of the divergence as a function ##r##, for ##\epsilon##<<1, ##\epsilon##≈1 , and ##\epsilon##>>1.
b. Calculate the flux of ##\vec{F}## outward through a sphere of radius R centered at the origin, and verify that it is equal to the integral of the divergence inside the sphere.
c. Show that the flux is ##4 \pi## (independent of R) in the limit ##\epsilon \rightarrow 0##.
Solution:
a. Calculating the divergence is simply a matter of plugging the function correctly into the spherical coordinates divergence formula. I get F_r=\frac{r}{(r^2+{\epsilon}^2)^{3/2}} \Rightarrow \frac{\partial F}{\partial r}=3r^2(r^2+{\epsilon}^2)^{-3/2}-3r^4(r^2+{\epsilon}^2)^{-5/2}
Therefore,
\nabla \cdot \vec{F}= 3(r^2+{\epsilon}^2)^{-3/2}-3r^2(r^2+{\epsilon}^2)^{-5/2}
So for the different values of ##\epsilon## do I just substitute?
For the vector function \vec{F}(\vec{r})=\frac{r\hat{r}}{(r^2+{\epsilon}^2)^{3/2}}
a. Calculate the divergence of ##\vec{F}(\vec{r})##, and sketch a plot of the divergence as a function ##r##, for ##\epsilon##<<1, ##\epsilon##≈1 , and ##\epsilon##>>1.
b. Calculate the flux of ##\vec{F}## outward through a sphere of radius R centered at the origin, and verify that it is equal to the integral of the divergence inside the sphere.
c. Show that the flux is ##4 \pi## (independent of R) in the limit ##\epsilon \rightarrow 0##.
Solution:
a. Calculating the divergence is simply a matter of plugging the function correctly into the spherical coordinates divergence formula. I get F_r=\frac{r}{(r^2+{\epsilon}^2)^{3/2}} \Rightarrow \frac{\partial F}{\partial r}=3r^2(r^2+{\epsilon}^2)^{-3/2}-3r^4(r^2+{\epsilon}^2)^{-5/2}
Therefore,
\nabla \cdot \vec{F}= 3(r^2+{\epsilon}^2)^{-3/2}-3r^2(r^2+{\epsilon}^2)^{-5/2}
So for the different values of ##\epsilon## do I just substitute?